## Section4.7Implicit and Logarithmic Differentiation

### Subsection4.7.1Implicit Differentiation

As we have seen, there is a close relationship between the derivatives of $\ds e^x$ and $\ln x$ because these functions are inverses. Rather than relying on pictures for our understanding, we would like to be able to exploit this relationship computationally. In fact this technique can help us find derivatives in many situations, not just when we seek the derivative of an inverse function.

We will begin by illustrating the technique to find what we already know, the derivative of $\ln x\text{.}$ Let's write $y=\ln x$ and then $\ds x=e^{\ln x}=e^y\text{,}$ that is, $\ds x=e^y\text{.}$ We say that this equation defines the function $y=\ln x$ implicitly because while it is not an explicit expression $y=\ldots\text{,}$ it is true that if $\ds x=e^y$ then $y$ is in fact the natural logarithm function. Now, for the time being, pretend that all we know of $y$ is that $\ds x=e^y\text{;}$ what can we say about derivatives? We can take the derivative of both sides of the equation:

\begin{equation*} \frac{d}{dx}x=\frac{d}{dx}e^y\text{.} \end{equation*}

Then using the Chain Rule on the right hand side:

\begin{equation*} 1 = \left(\frac{d}{dx}y\right) e^y = \frac{dy}{dx}e^y\text{.} \end{equation*}

Then we can solve for $\frac{dy}{dx}\text{:}$

\begin{equation*} \frac{dy}{dx}=\frac{1}{e^y} = \frac{1}{x}\text{.} \end{equation*}

There is one little difficulty here. To use the Chain Rule to compute $\ds d/dx(e^y)=\frac{dy}{dx}e^y$ we need to know that the function $y$ has a derivative. All we have shown is that if it has a derivative then that derivative must be $1/x\text{.}$ When using this method we will always have to assume that the desired derivative exists, but fortunately this is a safe assumption for most such problems.

The example $y=\ln x$ involved an inverse function defined implicitly, but other functions can be defined implicitly, and sometimes a single equation can be used to implicitly define more than one function.

###### Guideline for Implicit Differentiation.

Given an implicitly defined relation $f(x,y)=k$ for some constant $k\text{,}$ the following steps outline the implicit differentiation process for finding $dy/dx\text{:}$

1. Apply the differentiation operator $d/dx$ to both sides of the equation $f(x,y)=k\text{.}$

2. Follow through with the differentiation by keeping in mind that $y$ is a function of $x\text{,}$ and so the Chain Rule applies.

3. Solve for $dy/dx\text{.}$

Here's a familiar example.

###### Example4.68. Derivative of Circle Equation.

Given $r^2=x^2+y^2\text{,}$ find the derivative $\frac{dy}{dx}\text{.}$

Solution

The equation $r^{2}=x^{2}+y^{2}$ describes a circle of radius $r\text{.}$ The circle is not a function $y=f(x)$ because for some values of $x$ there are two corresponding values of $y\text{.}$ If we want to work with a function, we can break the circle into two pieces, the upper and lower semicircles, each of which is a function. Let's call these $y=U(x)$ and $y=L(x)\text{;}$ in fact this is a fairly simple example, and it's possible to give explicit expressions for these: $\ds U(x)=\sqrt{r^2-x^2\ }$ and $\ds L(x)=-\sqrt{r^2-x^2\ }\text{.}$ But it's somewhat easier, and quite useful, to view both functions as given implicitly by $\ds r^2=x^2+y^2\text{:}$ both $\ds r^2=x^2+U(x)^2$ and $\ds r^2=x^2+L(x)^2$ are true, and we can think of $\ds r^2=x^2+y^2$ as defining both $U(x)$ and $L(x)\text{.}$

Now we can take the derivative of both sides as before, remembering that $y$ is not simply a variable but a function—in this case, $y$ is either $U(x)$ or $L(x)$ but we're not yet specifying which one. When we take the derivative we just have to remember to apply the Chain Rule where $y$ appears.

\begin{equation*} \begin{split} \frac{d}{dx}r^2 =\amp \frac{d}{dx}(x^2+y^2)\\ 0 =\amp 2x+2y\frac{dy}{dx}\\ \frac{dy}{dx} =\amp \frac{-2x}{2y}=-\frac{x}{y}\end{split} \end{equation*}

Now we have an expression for $\frac{dy}{dx}\text{,}$ but it contains $y$ as well as $x\text{.}$ This means that if we want to compute $\frac{dy}{dx}$ for some particular value of $x$ we'll have to know or compute $y$ at that value of $x$ as well. It is at this point that we will need to know whether $y$ is $U(x)$ or $L(x)\text{.}$ Occasionally it will turn out that we can avoid explicit use of $U(x)$ or $L(x)$ by the nature of the problem.

###### Example4.69. Slope of the Circle.

Find the slope of the circle $\ds 4=x^2+y^2$ at the point $\ds (1,-\sqrt{3})\text{.}$

Solution

Since we know both the $x$- and $y$-coordinates of the point of interest, we do not need to explicitly recognize that this point is on $L(x)\text{,}$ and we do not need to use $L(x)$ to compute $y$ — but we could. Using the calculation of $\frac{dy}{dx}$ from above,

\begin{equation*} \frac{dy}{dx}=-\frac{x}{y}=-\frac{1}{-\sqrt{3}}=\frac{1}{\sqrt{3}}\text{.} \end{equation*}

It is instructive to compare this approach to others.

We might have recognized at the start that $\ds (1,-\sqrt{3})$ is on the function $\ds y=L(x)=-\sqrt{4-x^2}\text{.}$ We could then take the derivative of $L(x)\text{,}$ using the Power Rule and the Chain Rule, to get

\begin{equation*} L'(x)=-{1\over 2}(4-x^2)^{-1/2}(-2x)={x\over\sqrt{4-x^2}}\text{.} \end{equation*}

Then we could compute $\ds L'(1)=1/\sqrt{3}$ by substituting $x=1\text{.}$

Alternately, we could realize that the point is on $L(x)\text{,}$ but use the fact that $\frac{dy}{dx}=-x/y\text{.}$ Since the point is on $L(x)$ we can replace $y$ by $L(x)$ to get

\begin{equation*} \frac{dy}{dx}=-\frac{x}{L(x)}=-\frac{x}{\sqrt{4-x^2}}\text{,} \end{equation*}

without computing the derivative of $L(x)$ explicitly. Then we substitute $x=1$ and get the same answer as before.

In the case of the circle it is possible to find the functions $U(x)$ and $L(x)$ explicitly, but there are potential advantages to using implicit differentiation anyway. In some cases it is more difficult or impossible to find an explicit formula for $y$ and implicit differentiation is the only way to find the derivative.

###### Example4.70. Derivative of Function defined Implicitly.

Find the derivative of any function defined implicitly by $\ds yx^2+y^2=x\text{.}$

Solution

We treat $y$ as an unspecified function and use the Chain Rule:

\begin{equation*} \begin{gathered} \frac{d}{dx}(yx^2+y^2) = \frac{d}{dx}x\\ (y\cdot 2x+\frac{dy}{dx}\cdot x^2)+2y\frac{dy}{dx} = 1\\ \frac{dy}{dx}\cdot x^2+2y\frac{dy}{dx} = -y\cdot 2x\\ \frac{dy}{dx} =\frac{-2xy}{x^2+2y}\end{gathered} \end{equation*}
###### Example4.71. Derivative of Function defined Implicitly.

Find the derivative of any function defined implicitly by $\ds yx^2+e^y=x\text{.}$

Solution

We treat $y$ as an unspecified function and use the Chain Rule:

\begin{equation*} \begin{gathered} \frac{d}{dx}(yx^2+e^y) =\frac{d}{dx}x\\ (y\cdot 2x+\frac{dy}{dx}\cdot x^2)+\frac{dy}{dx}e^y = 1\\ \frac{dy}{dx}x^2+\frac{dy}{dx}e^y= 1-2xy\\ \frac{dy}{dx}(x^2+e^y)= 1-2xy\\ \frac{dy}{dx}=\frac{1-2xy}{x^2+e^y} \end{gathered} \end{equation*}

You might think that the step in which we solve for $\frac{dy}{dx}$ could sometimes be difficult—after all, we're using implicit differentiation here because we can't solve the equation $\ds yx^2+e^y=x$ for $y\text{,}$ so maybe after taking the derivative we get something that is hard to solve for $\frac{dy}{dx}\text{.}$ In fact, this never happens. All occurrences $\frac{dy}{dx}$ come from applying the Chain Rule, and whenever the Chain Rule is used it deposits a single $\frac{dy}{dx}$ multiplied by some other expression. So it will always be possible to group the terms containing $\frac{dy}{dx}$ together and factor out the $\frac{dy}{dx}\text{,}$ just as in the previous example. If you ever get anything more difficult you have made a mistake and should fix it before trying to continue.

It is sometimes the case that a situation leads naturally to an equation that defines a function implicitly.

###### Example4.72. Derivative of Function defined Implicitly.

Find $\ds\frac{dy}{dx}$ by implicit differentiation if

\begin{equation*} 2x^3+x^2y-y^9=3x+4\text{.} \end{equation*}
Solution

Differentiating both sides with respect to $x$ gives:

\begin{equation*} \begin{gathered} 6x^2+\left(2xy+x^2\frac{dy}{dx}\right)-9y^8\frac{dy}{dx}=3,\\ x^2\frac{dy}{dx}-9y^8\frac{dy}{dx}=3-6x^2-2xy\\ \left(x^2-9y^8\right)\frac{dy}{dx}=3-6x^2-2xy\\ \frac{dy}{dx}=\frac{3-6x^2-2xy}{x^2-9y^8}. \end{gathered} \end{equation*}
###### Example4.73. Derivative of Function defined Implicitly.

Suppose that $s$ and $t$ are related by the equation $s^{2}+te^{st}=2\text{.}$ Find $\frac{ds}{dt}\text{.}$

Solution

We assume that $s$ is a function of $t\text{,}$ $s(t)\text{.}$ Differentiate both sides of the equation defining the curve and group the terms involving $\frac{ds}{dt}$ obtaining,

\begin{equation*} \begin{gathered} \frac{d}{dt}\left(s^{2}+te^{st}\right) = \frac{d}{dt}2 \\ 2s\frac{ds}{dt} + e^{st} + te^{st}\left(s+t\frac{ds}{dt}\right) = 0 \\ \left(2s + t^2e^{st}\right)\frac{ds}{dt} + e^{st}(1+st) = 0 \end{gathered} \end{equation*}

We used the Product Rule and the Chain Rule to carry out the differentiation. Solving for $\frac{ds}{dt}$ gives

\begin{equation*} \frac{ds}{dt} = \frac{-e^{st}(1+st)}{2s+t^2e^{st}}=\frac{-(1+st)}{2se^{-st}+t^2} \cdot \end{equation*}

In the previous examples we had functions involving $x$ and $y\text{,}$ and we thought of $y$ as a function of $x\text{.}$ In these problems we differentiated with respect to $x\text{.}$ So when faced with $x$'s in the function we differentiated as usual, but when faced with $y$'s we differentiated as usual except we multiplied by a $\frac{dy}{dx}$ for that term because we were using Chain Rule.

### Subsection4.7.2Differentiating $x$ and $y$ as Functions of $t$

In the following example we will assume that both $x$ and $y$ are functions of $t$ and want to differentiate the equation with respect to $t\text{.}$ This means that every time we differentiate an $x$ we will be using the Chain Rule, so we must multiply by $\frac{dx}{dt}\text{,}$ and whenever we differentiate a $y$ we multiply by $\frac{dy}{dt}\text{.}$

###### Example4.74. Derivative of Function of an Additional Variable.

Thinking of $x$ and $y$ as functions of $t\text{,}$ i.e. $x(t)$ and $y(t)\text{,}$ differentiate the following equation with respect to $t\text{:}$

\begin{equation*} x^2+y^2=100\text{.} \end{equation*}
Solution

Using the Chain Rule we have:

\begin{equation*} 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\text{.} \end{equation*}
###### Example4.75. Derivative of Function of an Additional Variable.

If $y=x^3+5x$ and $\ds\frac{dx}{dt}=7\text{,}$ find $\ds\frac{dy}{dt}$ when $x=1\text{.}$

Solution

Differentiating each side of the equation $y=x^3+5x$ with respect to $t$ gives:

\begin{equation*} \frac{dy}{dt}=3x^2\frac{dx}{dt}+5\frac{dx}{dt}\text{.} \end{equation*}

When $x=1$ and $\frac{dx}{dt}=7$ we have:

\begin{equation*} \frac{dy}{dt}=3(1^2)(7)+5(7)=21+35=56\text{.} \end{equation*}
###### Example4.76. Differentiation with Parametric Functions.

Find $\frac{dy}{dx}$ when $\ds x = \cos t$ and $\ds y = \sin t\text{.}$

Solution

We differentiate both $x$ and $y$ with respect to the parameter $t\text{:}$

\begin{equation*} \frac{dx}{dt} = -\sin t \ \ \ \ \text{ and } \ \ \ \ \frac{dy}{dt} = \cos t \end{equation*}

From the Chain Rule, we know that

\begin{equation*} \frac{dy}{dx} = \frac{dy}{dx}\frac{dx}{dt} \end{equation*}

so that, by rearrangement

\begin{equation*} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\text{,} \end{equation*}

provided $\frac{dx}{dt} \neq 0\text{.}$ So, in this case

\begin{equation*} \frac{dy}{dx} = \frac{dy}{dx}\frac{dx}{dt} = \frac{\cos t}{\sin t} = -\cot t\text{.} \end{equation*}

### Subsection4.7.3Logarithmic Differentiation

Previously we've seen how to take the derivative of a number to a function $(a^{f(x)})'$ — exponential differentiation, and also a function to a number $[(f(x))^n]'$ — Power Rule. But what about the derivative of a function to a function $[(f(x))^{g(x)}]'\text{?}$

In this case, we use a procedure known as logarithmic differentiation.

###### Guideline for Logarithmic Differentiation.

Given a function $y=f(x)\text{,}$ the following steps outline the logarithmic differentiation process:

1. Take $\ln$ of both sides of $y=f(x)$ to get $\ln y=\ln f(x)$ and simplify using logarithm properties.

2. Differentiate implicitly with respect to $x$ and solve for $\ds\frac{dy}{dx}\text{.}$

3. Replace $y$ with its function of $x$ (i.e., $f(x)$).

###### Example4.77. Logarithmic Differentiation.

Differentiate $y=x^x\text{.}$

Solution

Method 1: We take $\ln$ of both sides:

\begin{equation*} \ln y=\ln x^x\text{.} \end{equation*}

Using log properties we have:

\begin{equation*} \ln y = x\ln x\text{.} \end{equation*}

Differentiating implicitly gives:

\begin{equation*} \frac{y'}{y}=(1)\ln x+x\frac{1}{x}\text{,} \end{equation*}
\begin{equation*} \frac{y'}{y}=\ln x+1\text{.} \end{equation*}

Solving for $y'$ gives:

\begin{equation*} y'=y(1+\ln x)\text{.} \end{equation*}

Replace $y=x^x$ gives:

\begin{equation*} y'=x^x(1+\ln x)\text{.} \end{equation*}

Method 2: Another method to find this derivative is as follows:

\begin{equation*} \begin{split} \frac{d}{dx}x^x =\amp \frac{d}{dx}e^{x\ln x}\\ =\amp \left(\frac{d}{dx}x\ln x\right)e^{x\ln x}\\ =\amp \left(x\frac{1}{x}+\ln x\right)x^x\\ =\amp (1+\ln x)x^x \end{split} \end{equation*}

In fact, logarithmic differentiation can be used on more complicated products and quotients (not just when dealing with functions to the power of functions).

###### Example4.78. Logarithmic Differentiation.

Differentiate (assuming $x>0$):

\begin{equation*} y=\frac{(x+2)^3(2x+1)^9}{x^8(3x+1)^4}\text{.} \end{equation*}
Solution

Using Product and Quotient Rules for this problem is a complete nightmare! Let's apply logarithmic differentiation instead. Take $\ln$ of both sides:

\begin{equation*} \ln y=\ln\left(\frac{(x+2)^3(2x+1)^9}{x^8(3x+1)^4}\right)\text{.} \end{equation*}

Applying log properties:

\begin{equation*} \ln y=\ln\left((x+2)^3(2x+1)^9\right)-\ln\left(x^8(3x+1)^4\right) \end{equation*}
\begin{equation*} \ln y=\ln\left((x+2)^3\right)+\ln\left((2x+1)^9\right)-\left[\ln\left(x^8\right)+\ln\left((3x+1)^4\right)\right] \end{equation*}
\begin{equation*} \ln y=3\ln(x+2)+9\ln(2x+1)-8\ln x-4\ln(3x+1)\text{.} \end{equation*}

Now, differentiating implicitly with respect to $x$ gives:

\begin{equation*} \frac{y'}{y}=\frac{3}{x+2}+\frac{18}{2x+1}-\frac{8}{x}-\frac{12}{3x+1}\text{.} \end{equation*}

Solving for $y'$ gives:

\begin{equation*} y'=y\left(\frac{3}{x+2}+\frac{18}{2x+1}-\frac{8}{x}-\frac{12}{3x+1}\right)\text{.} \end{equation*}

Replace $y=\frac{(x+2)^3(2x+1)^9}{x^8(3x+1)^4}$ gives:

\begin{equation*} y'=\frac{(x+2)^3(2x+1)^9}{x^8(3x+1)^4}\left(\frac{3}{x+2}+\frac{18}{2x+1}-\frac{8}{x}-\frac{12}{3x+1}\right)\text{.} \end{equation*}
###### Example4.79. Logarithmic Differentiation.

Differentiate $y = x^{2}(x-1)(x^{2}+4)^{3}\text{.}$

Solution

Taking the natural logarithm on both side of the given equation and using the laws of logarithms, we obtain

\begin{equation*} \begin{split} \ln y \amp = \ln\left(x^{2}(x-1)(x^{2}+4)^{3}\right)\\ \amp = \ln x^{2} + \ln(x-1) + \ln(x^{2}+4)^{3}\\ \amp = 2\ln x + \ln(x-1)+3\ln(x^{2}+4) \end{split} \end{equation*}

Differentiating both sides of the equation with respect to $x\text{,}$ we have

\begin{equation*} \frac{d}{dx} \ln y = \frac{y'}{y} = \frac{2}{x} + \frac{1}{x-1} + 3\frac{2x}{x^{2}+4} \cdot \end{equation*}

Finally, solving for $y'\text{,}$ we have

\begin{equation*} \begin{split} y' \amp = y\left(\frac{2}{x} + \frac{1}{x-1} + 3\frac{2x}{x^{2}+4}\right)\\ \amp = \left(x^{2}(x-1)(x^{2}+4)^{3}\right)\left(\frac{2}{x} + \frac{1}{x-1} + 3\frac{2x}{x^{2}+4}\right) \end{split} \end{equation*}
##### Exercises for Section 4.7.

Find a formula for the derivative $y'$ at the point $(x,y)\text{:}$

1. $\ds y^2=1+x^2$

$\ds x/y$
Solution
\begin{equation*} \diff{}{x}\left(y^{2}\right) = \diff{}{x}\left(1+x^{2}\right) \end{equation*}
\begin{equation*} 2yy'=2x \end{equation*}
\begin{equation*} y'(x) =\dfrac{x}{y} \end{equation*}
2. $\ds x^2+xy+y^2=7$

$\ds -(2x+y)/(x+2y)$
Solution
\begin{equation*} \diff{}{x} \left(x^2+xy+y^2\right) = \diff{}{x} (7) \end{equation*}
\begin{equation*} 2x + x \diff{y}{x} + y + 2 y \diff{y}{x} = 0 \end{equation*}
\begin{equation*} \diff{y}{x} \left(x + 2y\right) = -2x - y \end{equation*}
\begin{equation*} \diff{y}{x} = \dfrac{-2x-y}{x+2y} \end{equation*}
3. $\ds x^3+xy^2=y^3+yx^2$

$\ds (2xy-3x^2-y^2)/(2xy-3y^2-x^2)$
Solution
\begin{equation*} \diff{}{x} \left( x^3+xy^2\right) = \diff{}{x} \left(y^3+yx^2\right) \end{equation*}
\begin{equation*} 3x^2 + x(2y y') + y^2 = 3y^2y' + x^2 y' + 2xy \end{equation*}
\begin{equation*} y' (2xy -3y^2 - x^2) = -y^2-3x^2+2xy \end{equation*}
\begin{equation*} y'(x) = \dfrac{2xy-3x^2-y^2}{2xy-3y^2-x^2} \end{equation*}
4. $\ds 4\cos x \sin y = 1$

$\ds \sin(x)\sin(y)/(\cos(x)\cos(y))$
Solution
\begin{equation*} 4\diff{}{x}\left(\cos x \sin y\right) = \diff{}{x}\left(1\right) \end{equation*}
\begin{equation*} 4\sin x\sin y - 4\cos x \cos y y' = 0 \end{equation*}
\begin{equation*} y'(x) =\dfrac{\sin x \sin y}{\cos x \cos y} \end{equation*}
5. $\ds\sqrt{x} + \sqrt{y} = 9$

$\ds-\sqrt{y}/\sqrt{x}$
Solution
\begin{equation*} \diff{}{x} \left(\sqrt{x} + \sqrt{y} \right) = \diff{}{x} (9) \end{equation*}
\begin{equation*} \dfrac{1}{2\sqrt{x}} + \dfrac{y'}{2\sqrt{y}} = 0 \end{equation*}
\begin{equation*} y'(x) = \dfrac{-2\sqrt{y}}{2\sqrt{x}} \end{equation*}
\begin{equation*} y'(x) = \dfrac{-\sqrt{y}}{\sqrt{x}} \end{equation*}
6. $\ds \tan(x/y) = x+ y$

$\ds (y\sec^2(x/y)-y^2)/(x\sec^2(x/y)+y^2)$
Solution
\begin{equation*} \diff{}{x} \tan(x/y) = \diff{}{x} \left(x+y\right) \end{equation*}
\begin{equation*} \sec^2(x/y) \diff{}{x} \dfrac{x}{y} = 1 + y' \end{equation*}
\begin{equation*} \sec^2(x/y) \dfrac{y - xy'}{y^2} = 1 + y' \end{equation*}
\begin{equation*} y'(x) \left(-1 - \dfrac{x}{y^2}\sec^2(x/y)\right) = -\dfrac{\sec^2(x/y)}{y} \end{equation*}
\begin{equation*} y'(x) = \dfrac{\sec^2(xy)}{y(1+\dfrac{x}{y^2} \sec^2(xy)} \end{equation*}
\begin{equation*} y'(x) = \dfrac{y\sec^2(xy)-y^2}{x\sec^2(x/y) +y^2} \end{equation*}
7. $\ds \sin (x+y ) =xy$

$\ds (y-\cos(x+y))/(\cos(x+y)-x)$
Solution
\begin{equation*} \diff{}{x} \sin(x+y) = \diff{}{x} (xy) \end{equation*}
\begin{equation*} \cos(x+y)\left(1+y'\right) = y + xy' \end{equation*}
\begin{equation*} y'(x) \left(\cos(x+y)-x\right) = -\cos(x+y) + y \end{equation*}
\begin{equation*} y'(x) = \dfrac{y-\cos(x+y)}{\cos(x+y) - x} \end{equation*}
8. $\ds\frac{1}{x} + \frac{1}{y} = 7$

$\ds -y^2/x^2$
Solution
\begin{equation*} \diff{}{x} \left(\dfrac{1}{x} + \dfrac{1}{y} \right) = \diff{}{x} 7 \end{equation*}
\begin{equation*} -\dfrac{1}{x^2} - \dfrac{y'}{y^2} = 0 \end{equation*}
\begin{equation*} y'(x) = \dfrac{-y^2}{x^2} \end{equation*}
9. $\ds \cos(xy)-\sin x = 1$

$\ds -\frac{\cos x}{x\sin(xy)}-\frac{y}{x}$
Solution
\begin{equation*} \diff{}{x} \left(\cos(xy) - \sin x\right) = \diff{}{x} (1) \end{equation*}
\begin{equation*} -\sin(xy)\left(y + xy'\right) - \cos x = 0 \end{equation*}
\begin{equation*} -\sin(xy) x y' = \cos x +y\sin(xy) \end{equation*}
\begin{equation*} y'(x) = -\dfrac{\cos x + y \sin(xy)}{x\sin(xy)} \end{equation*}
10. $\ds x \sec(y)=\ln(\sin x)$

$\ds \frac{\cot x - \sec y}{x\sec y\tan y}$
Solution
\begin{equation*} \diff{}{x} x\sec y = \diff{}{x} \ln(\sin x) \end{equation*}
\begin{equation*} x \tan y \sec y y' + \sec y = \dfrac{\cos x}{\sin x} \end{equation*}
\begin{equation*} x \tan y \sec y y' = \dfrac{\cos x}{\sin x} - \sec y \end{equation*}
\begin{equation*} y'(x) = \dfrac{\cos x}{ x\sin x \tan y \sec y} - \dfrac{\sec y}{x\tan y \sec y} \end{equation*}
\begin{equation*} y'(x) = \dfrac{\cot x - \sec y}{x \tan y \sec y} \end{equation*}
11. $\ds \sin(x+y)-\sin^{-1}y = 0$

$\ds -\frac{\sqrt{1-y^{2}}\cos(x+y)}{\sqrt{1-y^{2}}\cos(x+y)-1}$
Solution
\begin{equation*} \diff{}{x} \left(\sin(x+y) - \sin^{-1} y\right) = \diff{}{x} (0) \end{equation*}
\begin{equation*} \cos(x+y) \left(1+y'\right) - \dfrac{y'}{1-y^2} = 0 \end{equation*}
\begin{equation*} \left(\cos(x+y)- \dfrac{1}{1-y^2}\right) y' = - \cos(x+y) \end{equation*}
\begin{equation*} y'(x) = -\dfrac{\sqrt{1-y^2}\cos(x+y)}{\sqrt{1-y^2}\cos(x+y) - 1} \end{equation*}
12. $\ds (x^{2}-y^{2})\tan(y) = \sqrt{y}$

$\ds \frac{4x\sqrt{y}\tan(y)}{4y^{3/2}\tan(y)-2\sqrt{y}(x^{2}-y^{2})\sec^{2}(y)+1}$
Solution
\begin{equation*} \diff{}{x} \left(x^2-y^2\right) \tan y = \diff{}{x} \sqrt{y} \end{equation*}
\begin{equation*} \left(2x - 2yy'\right)\tan y + \left(x^2-y^2\right) \sec^2y y' = \dfrac{y'}{2\sqrt{y}} \end{equation*}
\begin{equation*} y' \left(-2y\tan y +(x^2-y^2)\sec^2 y - \dfrac{1}{2\sqrt{y}}\right) = -2x \tan y \end{equation*}
\begin{equation*} y'(x) = \dfrac{-2x \tan y}{\left(-2y\tan y +(x^2-y^2)\sec^2 y - \dfrac{1}{2\sqrt{y}}\right)} \end{equation*}

Use logarithmic differentiation to find the derivative of $y\text{.}$

1. $\ds y=(x+1)^{2}(x+2)^{3}$

$\ds y'=(5x+7)(x+1)(x+2)^{2}$
Solution

We first take the logarithm of $y$ and simplify:

\begin{equation*} \ln y = \ln\left((x+1)^2(x+2)^3)\right) = 2\ln(x+1) + 3\ln(x+2)\text{.} \end{equation*}

Differentiating both sides, and solving for $y'\text{,}$ we find

\begin{equation*} \begin{split} \diff{}{x} \ln y \amp = \diff{}{x} \left(2\ln(x+1) + 3\ln(x+2)\right)\\ \frac{y'}{y} \amp = \frac{2}{x+1} + \frac{3}{x+2} \\ y'(x) \amp = y\left(\frac{2}{x+1} + \frac{3}{x+2}\right). \end{split} \end{equation*}

Therefore,

\begin{equation*} \begin{split} y'(x) \amp = (x+1)^2(x+2)^3\left(\frac{2}{x+1} + \frac{3}{x+2}\right)\\ \amp = 2(x+1)(x+2)^3 + 3(x+1)^2(x+2)^2\\ \amp = (5x+7)(x+1)(x+2)^2. \end{split} \end{equation*}
2. $\ds y=(3x+2)^{4}(5x-1)^{2}$

$\ds y'=2(3x+2)^{3}(5x-1)(45x+4)$
Solution

First, we take the logarithm of y. From the log rules,

\begin{equation*} \ln y = 4\ln\left(3x+2\right) + 2\ln\left(5x-1\right)\text{.} \end{equation*}

Now, we differentiate the right-hand side to get

\begin{equation*} \diff{}{x}\ln y = \frac{12}{3x+2} + \frac{10}{5x-1}\text{.} \end{equation*}

Finally, we note that the left-hand side, $\diff{}{x} \ln y$ is equal to $\dfrac{y'}{y}$ by the chain rule. Putting the two sides together and writing $y$ as a function of $x\text{,}$ we get

\begin{equation*} y'=y\left(\frac{12}{3x+2}+\frac{10}{5x-1}\right) = (3x+2)^{4}(5x-1)^{2}\left(\frac{12}{3x+2}+\frac{10}{5x-1}\right)\cdot \end{equation*}
3. $\ds y=(x-1)^{2}(x+1)^{3}(x+3)^{4}$

$\ds y'=(3x+2)^3(5x-1)(45x+4)$
Solution

First take the logarithm of both sides and simplify:

\begin{equation*} \begin{split} \ln(y) \amp = \ln\left((x-1)^2(x+1)^3(x+3)^4\right)\\ \amp =2\ln(x-1) + 3\ln(x+2) + 4\ln(x+3) \end{split} \end{equation*}

Now differentiate both sides and solve for $y'\text{:}$

\begin{equation*} \begin{split} \diff{}{x} \ln(y) \amp = \diff{}{x} \left(2\ln(x-1) + 3\ln(x+2) + 4\ln(x+3) \right)\\ \frac{y'}{y} \amp = \frac{2}{x-1} + \frac{3}{x+2} + \frac{4}{x+3} \\ y'(x) \amp = y\left(\frac{2}{x-1} + \frac{3}{x+2} + \frac{4}{x+3}\right)\\ y'(x) \amp = (x-1)^2(x+1)^3(x+3)^4\left(\frac{2}{x-1} + \frac{3}{x+2} + \frac{4}{x+3}\right)\\ y'(x) \amp = 2(3x+2)^3(5x-1)(45x+4) \end{split} \end{equation*}
4. $\ds y=\sqrt{3x+5}(2x-3)^{4}$

$\ds y'=\sqrt{3x+5}(2x-3)^4\left( \frac{3}{2(3x+5)} + \frac{8}{2x-3}\right)$
Solution

We first find $\ln(y)\text{:}$

\begin{equation*} \begin{split} \ln(y) \amp = \ln\left(\sqrt{3x+5}(2x-3)^4\right)\\ \amp = \frac{1}{2}\ln(3x+5) + 4\ln(2x-3) \end{split} \end{equation*}

Now differentiate, applying the Chain Rule as necessary:

\begin{equation*} \begin{split} \diff{}{x} \ln(y) \amp = \diff{}{x} \left( \frac{1}{2}\ln(3x+5) + 4\ln(2x-3) \right)\\ \frac{y'}{y} \amp = \frac{1}{2(3x+5)}\diff{}{x}(3x+5) + \frac{4}{2x-3} \diff{}{x} (2x-3) \\ \frac{y'}{y} \amp = \frac{3}{2(3x+5)} + \frac{4(2)}{2x-3}\\ y'(x) \amp = y\left( \frac{3}{2(3x+5)} + \frac{8}{2x-3}\right)\\ y'(x) \amp = \sqrt{3x+5}(2x-3)^4\left( \frac{3}{2(3x+5)} + \frac{8}{2x-3}\right) \end{split} \end{equation*}
5. $\ds y=\dfrac{(2x^{2}-1)^{5}}{\sqrt{x+1}}$

$\ds y'=\frac{(2x^2-1)^5}{\sqrt{x+1}} \left(\frac{20x^2}{2x^2-1} - \frac{1}{2(x+1)}\right)$
Solution

We take the logarithm of $y$ to find

\begin{equation*} \begin{split} \ln(y) \amp = \ln \left(\frac{(2x^2-1)^5}{\sqrt{x+1}}\right) \\ \amp = 5\ln(2x^2-1) - \frac{1}{2} \ln(x+1) \end{split} \end{equation*}

Differentiating both sides of the above equation gives

\begin{equation*} \begin{split} \diff{}{x} \ln(y) \amp = \diff{}{x} \left(5\ln(2x^2-1) - \frac{1}{2} \ln(x+1)\right) \\ \frac{y'}{y} \amp = \frac{5}{2x^2-1} \diff{}{x} (2x^2-1) - \frac{1}{2(x+1)} \diff{}{x} (x+1) \\ \frac{y'}{y} \amp = \frac{20x^2}{2x^2-1} - \frac{1}{2(x+1)} \end{split} \end{equation*}

Therefore, we can write $y'(x)$ explicitly:

\begin{equation*} y'(x) =\frac{(2x^2-1)^5}{\sqrt{x+1}} \left(\frac{20x^2}{2x^2-1} - \frac{1}{2(x+1)}\right) \end{equation*}
6. $\ds y=\dfrac{\sqrt{4+3x^{2}}}{\sqrt[3]{x^{2}+1}}$

$y'(x) = \frac{(4+3x^2)^{1/2}}{(x^2+1)^{1/3}}\left(\frac{3x}{4+3x^2} - \frac{2x}{3(x^2+1)}\right)$
Solution

Write

\begin{equation*} y = \frac{(4+3x^2)^{1/2}}{(x^2+1)^{1/3}}\text{.} \end{equation*}

Then, we find $\ln(y)\text{:}$

\begin{equation*} \ln(y) = \frac{1}{2} \ln(4+3x^2) - \frac{1}{3} \ln(x^2+1) \end{equation*}

Hence,

\begin{equation*} \begin{split} \diff{}{x} \ln (y) \amp = \diff{}{x} \left(\frac{1}{2} \ln(4+3x^2) - \frac{1}{3} \ln(x^2+1) \right)\\ \frac{y'}{y} \amp = \frac{1}{2(4+3x^2)} \diff{}{x}(4+3x^2) - \frac{1}{3(x^2+1)}\diff{}{x}(x^2+1) \\ \frac{y'}{y}\amp = \frac{1}{2(4+3x^2)} (6x) - \frac{1}{3(x^2+1)} (2x) \\ y'(x) \amp = y\left(\frac{3x}{4+3x^2} - \frac{2x}{3(x^2+1)}\right). \end{split} \end{equation*}

Therefore,

\begin{equation*} y'(x) = \frac{(4+3x^2)^{1/2}}{(x^2+1)^{1/3}}\left(\frac{3x}{4+3x^2} - \frac{2x}{3(x^2+1)}\right)\text{.} \end{equation*}
7. $\ds y=3^{x}$

$\ds y'=3^{x}\ln 3$
Solution

Notice that the rule for exponential functions of base 3 says that $\diff{}{x}3^{x}=3^{x}\ln 3\text{.}$ Let's verify using logarithmic differentiation.

\begin{equation*} \begin{split} \amp \ln y = x \ln 3 \amp \diff{}{x}\left(\ln y\right) = \ln 3 \amp y' = y \ln 3 = 3^{x}\ln 3, \end{split} \end{equation*}

as expected.

8. $\ds y=x^{x+2}$

$\ds y'= x^{x+2}\left(\ln(x) + \frac{x+2}{x} \right)$
Solution

Taking the logarithm of $y$ gives

\begin{equation*} \ln(y) = (x+2)\ln(x)\text{.} \end{equation*}

Now take the derivative of both sides:

\begin{equation*} \begin{split} \diff{}{x} \ln(y) \amp = \diff{}{x} (x+2)\ln(x)\\ \frac{y'}{y} \amp = \ln(x) \diff{}{x} (x+2) + (x+2) \diff{}{x} \ln(x) \\ \frac{y'}{y} \amp = \ln(x) + \frac{x+2}{x} \\ y' \amp = y\left(\ln(x) + \frac{x+2}{x} \right) \end{split} \end{equation*}

Hence,

\begin{equation*} y'(x) = x^{x+2}\left(\ln(x) + \frac{x+2}{x} \right)\text{.} \end{equation*}
9. $\ds y=(x^{2}+1)^{x}$

$\ds y'(x) = x \ln(x^2+1)\left( \frac{2x^2}{x^2+1} + \ln(x^2+1)\right)$
Solution

We have

\begin{equation*} \ln(y) = x \ln(x^2+1)\text{.} \end{equation*}

Therefore,

\begin{equation*} \begin{split} \diff{}{x} \ln(y) \amp = \diff{}{x} x \ln(x^2+1) \\ \frac{y'}{y} \amp = x\diff{}{x} \ln(x^2+1) + \ln(x^2+1)\diff{}{x} (x)\\ \frac{y'}{y} \amp = \frac{x}{x^2+1} \diff{}{x} (x^2+1) + \ln(x^2+1)(1) \\ \frac{y'}{y} \amp = \frac{x}{x^2+1} (2x) + \ln(x^2+1)\\ y' \amp = y\left( \frac{2x^2}{x^2+1} + \ln(x^2+1)\right) \end{split} \end{equation*}

Hence, we can write $y'$ explicitly:

\begin{equation*} y'(x) = x \ln(x^2+1)\left( \frac{2x^2}{x^2+1} + \ln(x^2+1)\right) \end{equation*}
10. $\ds y=x^{\ln x}$

$\ds x^{\ln x}\left(\frac{2\ln(x)}{x}\right)$
Solution

First, find $\ln(y)\text{:}$

\begin{equation*} \ln(y) = \ln(x^{\ln x}) = \ln(x)\ln(x) = \left(\ln x\right)^2 \end{equation*}

Therefore, differentiating and applying the Chain Rule gives

\begin{equation*} \begin{split} \diff{}{x} \ln(y) \amp = \diff{}{x} \left(\ln x\right)^2 \\ \frac{y'}{y} \amp = 2\ln(x) \diff{}{x} \ln(x) \\ y' \amp = y\left(\frac{2\ln(x)}{x}\right) \end{split} \end{equation*}

Hence,

\begin{equation*} y'(x) = x^{\ln x}\left(\frac{2\ln(x)}{x}\right)\text{.} \end{equation*}

A hyperbola passing through $(8,6)$ consists of all points whose distance from the origin is a constant more than its distance from the point (5,2). Find the slope of the tangent line to the hyperbola at $(8,6)\text{.}$

$1$
Solution

We first find an equation of the curve. The distance from the point $(8,6)$ to the origin is

\begin{equation*} \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{.} \end{equation*}

Furthermore, the from the point $(8,6)$ to the point $(5,2)$ is

\begin{equation*} \sqrt{(6-2)^2 + (8-5)^2} = \sqrt{25} = 5\text{.} \end{equation*}

And so we must have

\begin{equation*} \sqrt{x^2+y^2} - \sqrt{(x-5)^2 + (y-2)^2} = 5 \end{equation*}

for all points $(x,y)$ on the curve. We now differentiate implicitly:

\begin{equation*} \begin{split} \diff{}{x} \left(\sqrt{x^2+y^2} - \sqrt{(x-5)^2 + (y-2)^2} \right) \amp = \diff{}{x} (5)\\ \frac{2x + 2yy'}{2\sqrt{x^2+y^2}} - \frac{2(x-5) + 2(y-2) y'}{2\sqrt{(x-5)^2 + (y-2)^2}} \amp = 0\\ -\frac{\frac{x}{\sqrt{x^2+y^2}} - \frac{(x-5)}{\sqrt{(x-5)^2 + (y-2)^2}}}{\left(\frac{(y-2)}{\sqrt{(x-5)^2 + (y-2)^2}} - \frac{y}{\sqrt{x^2+y^2}}\right)} \amp = y' \end{split} \end{equation*}

Hence, at the point $(8,6)\text{,}$ the slope of the tangent line must be

\begin{equation*} \begin{split} y'\amp= -\frac{\frac{8}{\sqrt{8^2+6^2}} - \frac{(8-5)}{\sqrt{(8-5)^2 + (6-2)^2}}}{\left(\frac{(6-2)}{\sqrt{(8-5)^2 + (6-2)^2}} - \frac{6}{\sqrt{8^2+6^2}}\right)}\\\amp= 1.\end{split} \end{equation*}

The graph of the equation $\ds x^2 - xy + y^2 = 9$ is an ellipse. Find the lines tangent to this curve at the two points where it intersects the $x$-axis. Show that these lines are parallel.

$y=2x\pm6$
Solution

Let $y=0\text{.}$ Then

\begin{equation*} x^2-x(0)+(0)^2 = 9 \implies x^2 = 9 \implies x = \pm 3\text{.} \end{equation*}

And so the ellipse intersects the $x$-axis at the points $(3,0)$ and $(-3,0)\text{.}$ To find the slope of the tangent lines at each of the two points, we differentiate the curve implicitly:

\begin{equation*} \begin{gathered} \diff{}{x} \left(x^2-xy+y^2\right) = \diff{}{x} 9 \\ 2x - (y + xy') + 2yy' = 0 \\ 2yy'-xy' = -2x+y \\ y' = \frac{-2x+y}{2y-x} \end{gathered} \end{equation*}

Hence, at the point $(3,0)$ the slope of the tangent line is

\begin{equation*} \frac{-2x+y}{2y-x} \bigg\vert_{x=3,y=0} = \frac{-6}{-3} = 2\text{.} \end{equation*}

Therefore, an equation of the tangent line is

\begin{equation*} y = 2(x-3)\text{.} \end{equation*}

At the point $(-3,0)$ the slope of the tangent line is

\begin{equation*} \frac{-2x+y}{2y-x} \bigg\vert_{x=-3,y=0} = \frac{6}{2} = 2\text{.} \end{equation*}

Therefore, an equation of the tangent line is

\begin{equation*} y = 2(x+3)\text{.} \end{equation*}

The tangent lines are parallel since they have the same slope.

Repeat the previous problem for the points at which the ellipse intersects the $y$-axis.

$y=x/2\pm3$
Solution

Let $x=0\text{.}$ Then

\begin{equation*} 0^2-(0)y+y^2 = 9 \implies y^2 = 9 \implies y = \pm 3\text{.} \end{equation*}

And so the ellipse intersects the $y$-axis at the points $(0,3)$ and $(0,-3)\text{.}$ To find the slope of the tangent lines at each of the two points, we differentiate the curve implicitly (as before):

\begin{equation*} \begin{gathered} \diff{}{x} \left(x^2-xy+y^2\right) = \diff{}{x} 9 \\ 2x - (y + xy') + 2yy' = 0 \\ 2yy'-xy' = -2x+y \\ y' = \frac{-2x+y}{2y-x} \end{gathered} \end{equation*}

Hence, at the point $(0,3)$ the slope of the tangent line is

\begin{equation*} \frac{-2x+y}{2y-x} \bigg\vert_{x=0,y=3} = \frac{3}{6} = \frac{1}{2}\text{.} \end{equation*}

Therefore, an equation of the tangent line is

\begin{equation*} y-3 = \frac{x}{2}\text{.} \end{equation*}

At the point $(-3,0)$ the slope of the tangent line is

\begin{equation*} \frac{-2x+y}{2y-x} \bigg\vert_{x=0,y=-3} = \frac{-3}{-6} = \frac{1}{2}\text{.} \end{equation*}

Therefore, an equation of the tangent line is

\begin{equation*} y+3 = \frac{x}{2}\text{.} \end{equation*}

The tangent lines are parallel since they have the same slope.

If $\ds y=\log_a x$ then $\ds a^y=x\text{.}$ Use implicit differentiation to find $\ds y'\text{.}$

$y'(x) = = \frac{1}{x\ln a}$
Solution

We differentiate:

\begin{equation*} \begin{split} \diff{}{x} a^{y} \amp = \diff{}{x} x \\ a^y \ln(a) y' \amp = 1 \\ y' \amp = \frac{1}{\ln a} a^{-y} \end{split} \end{equation*}

Since $y=\log_a{x}\text{,}$ we write

\begin{equation*} y'(x) = \frac{1}{\ln a} a^{-\log_a x} =\frac{1}{\ln a} a^{\log_a x^{-1}} = \frac{1}{x\ln a}\text{.} \end{equation*}

Find the points on the ellipse from the previous two problems where the slope is horizontal and where it is vertical.

$\ds (\sqrt3,2\sqrt3)\text{,}$ $\ds (-\sqrt3,-2\sqrt3)\text{,}$ $\ds (2\sqrt3,\sqrt3)\text{,}$ $\ds (-2\sqrt3,-\sqrt3)$
Solution

We wish to finds the points on the ellipse $x^2-xy+y^2=9$ such that the tangent is horizontal and where the tangent is vertical. As before, we differentiate implicitly:

\begin{equation*} \begin{gathered} \diff{}{x} \left(x^2-xy+y^2\right) = \diff{}{x} 9 \\ 2x - (y + xy') + 2yy' = 0 \\ 2yy'-xy' = -2x+y \\ y' = \frac{-2x+y}{2y-x} \end{gathered} \end{equation*}

To find where the tangent line is horizontal, we set

\begin{equation*} -2x + y = 0 \implies y=2x\text{.} \end{equation*}

Thus, we need to find where the line $y=2x$ intersects the ellipse:

\begin{equation*} x^2 - x(2x) + (2x)^2 = 9 \implies 3x^2 = 9 \implies x = \pm \sqrt{3}\text{.} \end{equation*}

Hence, the tangent line is horizontal at the points $(\sqrt{3}, 2\sqrt{3})$ and $(-\sqrt{3},-2\sqrt{3})\text{.}$

To find where the tangent is vertical, we set

\begin{equation*} 2y-x =0 \implies x = 2y\text{.} \end{equation*}

Thus, we need to find the two points where the line $x=2y$ intersects the ellipse:

\begin{equation*} (2y)^2-(2y)y + y^2 = 9 \implies 3y^2 = 9 \implies y = \pm \sqrt{3}\text{.} \end{equation*}

Hence, the tangent line is vertical at the points $(\sqrt{3},2\sqrt{3})$ and $(-\sqrt{3},-2\sqrt{3})\text{.}$

Find an equation for the tangent line to $\ds x^4 = y^2 + x^2$ at $\ds (2, \sqrt{12})\text{.}$ (This curve is the kampyle of Eudoxus.)

$\ds y=7x/\sqrt3-8/\sqrt3$
Solution

We find the slope of the tangent line at any point $\left(x,y\right)$ by differentiating the curve implicitly.

\begin{equation*} \begin{split} \amp \diff{}{x}\left(x^{4}\right) = \diff{}{x} \left(y^{2}+x^{2}\right) \amp 4x^{3} = 2yy'+2x \amp y' = \frac{4x^{3}-2x}{2y} \end{split} \end{equation*}

Therefore, at the point $\left(2,\sqrt{12}\right)\text{,}$ $y' = \frac{4(8)-2(2)}{2\sqrt{12}} = \frac{7}{\sqrt{3}}\text{.}$ And so the equation of the tangent line to the curve at this point is

\begin{equation*} y = \frac{7}{\sqrt{3}} \left(x-2\right) + \sqrt{12} = \frac{7}{\sqrt{3}}x - \frac{8}{\sqrt{3}}\text{.} \end{equation*}

Find an equation for the tangent line to $\ds x^{2/3} + y^{2/3} = a^{2/3}$ at a point $\ds (x_1 ,y_1)$ on the curve, with $\ds x_1 \neq 0$ and $\ds y_1 \neq 0\text{.}$ (This curve is an astroid.)

$\ds y=(-y_1^{1/3}x+y_1^{1/3}x_1+x_1^{1/3}y_1)/x_1^{1/3}$
Solution

Differentiate the curve implicitly:

\begin{equation*} \begin{gathered} \diff{}{x} \left(x^{2/3} + y^{2/3}\right) = \diff{}{x} a^{2/3} \\ \frac{2}{3} x^{1/3} + \frac{2}{3}y^{1/3}y' = 0 \\ y' = -\frac{x^{1/3}}{y^{1/3}} \end{gathered} \end{equation*}

So the slope $m$ of the tangent line at the point $(x_1,y_1)$ is

\begin{equation*} m = -\frac{x_1^{1/3}}{y_1^{1/3}}\text{,} \end{equation*}

for $x_1, y_1 \neq 0\text{.}$

Using the point-slope formula, we find the equation of the tangent line to be

\begin{equation*} -\frac{x_1^{1/3}}{y_1^{1/3}} = \frac{y-y_1}{x-x_1}\text{.} \end{equation*}

Find an equation for the tangent line to $\ds (x^2 +y^2 )^2 =x^2 -y^2$ at a point $\ds (x_1 , y_1)$ on the curve, with $\ds x_1 \neq 0, -1, 1\text{.}$ (This curve is a lemniscate.)

$\ds (y-y_1)/(x-x_1)=(2x_1^3+2x_1y_1^2-x_1)/(2y_1^3+2y_1x_1^2+y_1)$
Solution

We differentiate implicitly:

\begin{equation*} \begin{split} \diff{}{x} (x^2+y^2)^2 \amp = \diff{}{x} (x^2-y^2) \\ 2(x^2+y^2) \diff{}{x} (x^2+y^2) \amp = 2x - 2y y' \\ (x^2+y^2) (2x+2yy') \amp = x - y y' \\ 2yy'(x^2+y^2)+yy' \amp = x - 2x(x^2+y^2)\\ y' \amp = \frac{-2x^3+x-2xy^2}{2x^2+2y^3+y} \end{split} \end{equation*}

Therefore, the slope of the tangent line at the point $(x_1,y_1)$ is

\begin{equation*} \frac{-2x_1^3+x_1-2x_1y_1^2}{2x_1^2+2y_1^3+y_1}\text{,} \end{equation*}

and so an equation of the tangent line is

\begin{equation*} y - y_1 = \frac{-2x_1^3+x_1-2x_1y_1^2}{2x_1^2+2y_1^3+y_1}\left(x-x_1\right)\text{.} \end{equation*}

Two curves are orthogonal if at each point of intersection, the angle between their tangent lines is $\pi/2\text{.}$ Two families of curves, $\cal{A}$ and $\cal{B}\text{,}$ are orthogonal trajectories of each other if given any curve $C$ in $\cal{A}$ and any curve $D$ in $\cal{B}$ the curves $C$ and $D$ are orthogonal. For example, the family of horizontal lines in the plane is orthogonal to the family of vertical lines in the plane.

1. Show that $\ds x^2 -y^2 =5$ is orthogonal to $\ds 4x^2 +9y^2 =72\text{.}$ Hint

You need to find the intersection points of the two curves and then show that the product of the derivatives at each intersection point is $-1\text{.}$

Solution

Notice that the first equation is a hyperbola and the second equation is an ellipse. Let's first differentiate each curve implicitly:

\begin{equation*} \begin{split} \diff{}{x} \left(x^2-y^2\right) \amp = \diff{}{x} 5 \\ 2x - 2yy' \amp = 0 \\ y' \amp = \frac{x}{y} \end{split} \end{equation*}

This gives the slope of the hyperbola. Next,

\begin{equation*} \begin{split} \diff{}{x} \left(4x^2+9y^2\right) \amp = \diff{}{x} 72\\ 8x + 18yy' \amp = 0 \\ y' \amp = -\frac{4x}{9y} \end{split} \end{equation*}

This gives the slope of the ellipse. We now find the intersection points. From the equation, let $x^2=5+y^2\text{.}$ Then, from the second equation, we have

\begin{equation*} \begin{split} 4(5+y^2) + 9y^2 \amp = 72\\ 20 + 4y^2 + 9y^2 \amp = 72\\ 13y^2 \amp = 52 \\ y \amp = \pm \sqrt{4} = \pm 2. \end{split} \end{equation*}

When $y^2=4\text{,}$ we have $x^2 = 9\text{.}$ Hence, there are four intersection points: $(3,2)\text{,}$ $(3,-2)\text{,}$ $(-3,2)$ and $(-3,-2)\text{.}$ Therefore, at the point $(2,3)\text{,}$ the product of the slopes is

\begin{equation*} -\frac{4(3)}{9(2)} \cdot \frac{3}{2} = -1\text{,} \end{equation*}

and similarly for all other intersection points. Hence, the curves are orthogonal.

2. Show that $\ds x^2 +y^2 = r^2$ is orthogonal to $y=mx\text{.}$ Conclude that the family of circles centered at the origin is an orthogonal trajectory of the family of lines that pass through the origin. Note that there is a technical issue when $m=0\text{.}$ The circles fail to be differentiable when they cross the $x$-axis. However, the circles are orthogonal to the $x$-axis. Explain why. Likewise, the vertical line through the origin requires a separate argument.

Solution

We are given an equation of a circle centred at the origin and the equation of a line which passes through the origin. The slope of the circle at any point $(x,y)$ is given by

\begin{equation*} 2x + 2yy' = 0 \implies y' = -\frac{x}{y}\text{,} \end{equation*}

and the slope of the line is always $m\text{.}$ The lines will intersect with the circle at exactly two points. At both intersection points, we require $y=mx$ and so the product of the slopes of the two curves is

\begin{equation*} m \cdot \frac{-x}{mx} = -1\text{.} \end{equation*}

Hence, the family of circles centred at the original is an orthogonal tranjectory of the family of lines which pass through the origin. When $m=0\text{,}$ the line lies along the $x$-axis. Since the tangent line to the circle is vertical when it intersects with the $x$-axis, the curves remain orthogonal. Similarly, the tangent line to the circle is horizontal when it intersects the $y$-axis, and so again the curves remain orthogonal.

3. For $k\not= 0$ and $c \neq 0$ show that $\ds y^2 -x^2 =k$ is orthogonal to $yx =c\text{.}$ In the case where $k$ and $c$ are both zero, the curves intersect at the origin. Are the curves $\ds y^2 -x^2 =0$ and $yx=0$ orthogonal to each other?

No.
Solution

The slope of the first hyperbola is found to be

\begin{equation*} 2yy' - 2x = 0 \implies y' = \frac{x}{y}\text{,} \end{equation*}

and the slope of the second hyperbola is

\begin{equation*} y' = \frac{-c}{x^2}\text{.} \end{equation*}

Since we require that $yx = c$ (for $c\neq 0$), this means that the product of the slopes will be

\begin{equation*} \frac{x^2}{c} \cdot \frac{-c}{x^2} = -1\text{.} \end{equation*}

Hence, for $k\neq 0$ and $c\neq 0\text{,}$ the curves are orthogonal trajectories. The equation $yx = 0$ is satisfied along the $y$-axis and along the $x$-axis. However, the equation $x^2=y^2$ is satisfied along the lines $y=x$ and $y=-x\text{.}$ Therefore, the angle between the curves $xy=0$ and $x^2-y^2=0$ is $\pi/4\text{,}$ and so they are not orthogonal.

4. Suppose that $m\neq 0\text{.}$ Show that the family of curves $\ds \{y=mx+b \mid b\in \R \}$ is orthogonal to the family of curves $\ds \{y=-(x/m)+c \mid c \in \R\}\text{.}$

Differentiate the function $\ds y={(x-1)^8 (x-23)^{1/2}\over 27 x^6(4x-6)^8 }$

Solution

We use logarithmic differentiation. First, find $\ln y\text{:}$

\begin{equation*} \begin{split} \ln y \amp = \ln \left(\frac{(x-1)^8(x-23)^{1/2}}{27x^6(4x-6)^8}\right)\\ \amp = \ln\bigl((x-1)^8(x-23)^{1/2}\bigr)-\ln\bigl((27x^6(4x-6)^8\bigr)\\ \amp = \left(8\ln(x-1) + \frac{1}{2}\ln(x-3)\right) - \left(162\ln(x) + 8\ln(4x-6)\right) \end{split} \end{equation*}

We now differentiate both sides of the above equation:

\begin{equation*} \begin{split} \diff{}{x} \ln y \amp = \diff{}{x} \left(8\ln(x-1) + \frac{1}{2}\ln(x-3)\right) - \left(162\ln(x) + 8\ln(4x-6)\right) \\ \frac{y'}{y} \amp = \frac{8}{x-1} + \frac{1}{2(x-3)} - \frac{162}{x} - \frac{8(4)}{4x-6} \end{split} \end{equation*}

Therefore,

\begin{equation*} y'(x) = \left(\frac{(x-1)^8(x-23)^{1/2}}{27x^6(4x-6)^8}\right)\left(\frac{8}{x-1} + \frac{1}{2(x-3)} - \frac{162}{x} - \frac{8(4)}{4x-6} \right)\text{.} \end{equation*}

Differentiate the function $\ds f(x)=(x+1)^{\sin x}\text{.}$

Solution

We use logarithmic differentiation. First find $\ln f(x)\text{:}$

\begin{equation*} \ln (f(x)) = \sin (x) \ln(x+1)\text{.} \end{equation*}

Therefore,

\begin{equation*} \begin{split} \diff{}{x} \ln(f(x)) \amp = \diff{}{x} \sin(x) \ln(x+1) \\ \frac{f'(x)}{f(x)} \amp = \cos(x) \ln(x+1) + \frac{\sin (x)}{x+1} \\ f'(x) \amp = f(x) \left(\cos(x) \ln(x+1) + \frac{\sin (x)}{x+1}\right) \end{split} \end{equation*}

Hence, written explicitly as a function of $x\text{,}$ we have

\begin{equation*} f'(x) = \sin (x) \ln(x+1) \left(\cos(x) \ln(x+1) + \frac{\sin (x)}{x+1}\right)\text{.} \end{equation*}

Differentiate the function $\ds g(x)=\frac{e^x(\cos x+2)^3}{\sqrt{x^2+4}}\text{.}$

Solution

We use logarithmic differentiation. First find $\ln(g(x))\text{:}$

\begin{equation*} \ln(g(x)) = \ln(e^x) + 3\ln(\cos(x)+2) - \frac{1}{2} \ln(x^2+4) \end{equation*}

Thus,

\begin{equation*} \begin{split} \diff{}{x} \ln(g(x)) \amp = \diff{}{x} \left(\ln(e^x) + 3\ln(\cos(x)+2) - \frac{1}{2} \ln(x^2+4)\right) \\ \frac{g'(x)}{g(x)} \amp = \frac{1}{e^x} + \frac{3}{\cos(x) +2} \left(-\sin(x)\right) - \frac{1}{2(x^2+4)}\left(2x\right) \\ g'(x) \amp = g(x) \left(\frac{1}{e^x} + \frac{-3\sin(x)}{\cos(x) +2}- \frac{x}{x^2+4}\right) \end{split} \end{equation*}

Therefore, we have found that

\begin{equation*} g'(x) = \frac{e^x(\cos x + 2)^3}{\sqrt{x^2+4}} \left(\frac{1}{e^x} + \frac{-3\sin(x)}{\cos(x) +2}- \frac{x}{x^2+4}\right) \end{equation*}