## Section4.6Derivatives of Exponential & Logarithmic Functions

As with the sine function, we don't know anything about derivatives that allows us to compute the derivatives of the exponential and logarithmic functions without going back to basics. Let's do a little work with the definition again:

\begin{equation*} \begin{split} \frac{d}{dx}a^x =\amp \lim_{\Delta x\to 0} \frac{a^{x+\Delta x}-a^x}{\Delta x}\\ =\amp \lim_{\Delta x\to 0} \frac{a^xa^{\Delta x}-a^x}{\Delta x}\\ =\amp \lim_{\Delta x\to 0} a^x\frac{a^{\Delta x}-1}{\Delta x} \\ =\amp a^x\lim_{\Delta x\to 0} \frac{a^{\Delta x}-1}{\Delta x}\end{split} \end{equation*}

There are two interesting things to note here: As in the case of the sine function we are left with a limit that involves $\Delta x$ but not $x\text{,}$ which means that if $\ds \lim_{\Delta x\to 0} (a^{\Delta x}-1)/\Delta x$ exists, then it is a constant number. This means that $\ds a^x$ has a remarkable property: its derivative is a constant times itself.

We earlier remarked that the hardest limit we would compute is $\ds \lim_{x\to0}\sin x/x=1\text{;}$ we now have a limit that is just a bit too hard to include here. In fact the hard part is to see that $\ds \lim_{\Delta x\to 0} (a^{\Delta x}-1)/\Delta x$ even exists—does this fraction really get closer and closer to some fixed value? Yes it does, but we will prove this property at the end of this section.

We can look at some examples. Consider $\ds (2^x-1)/x$ for some small values of $x\text{:}$ 1, $0.828427124\text{,}$ $0.756828460\text{,}$ $0.724061864\text{,}$ $0.70838051\text{,}$ $0.70070877$ when $x$ is 1, $1/2\text{,}$ $1/4\text{,}$ $1/8\text{,}$ $1/16\text{,}$ $1/32\text{,}$ respectively. It looks like this is settling in around $0.7\text{,}$ which turns out to be true (but the limit is not exactly $0.7$). Consider next $\ds (3^x-1)/x\text{:}$ $2\text{,}$ $1.464101616\text{,}$ $1.264296052\text{,}$ $1.177621520\text{,}$ $1.13720773\text{,}$ $1.11768854\text{,}$ at the same values of $x\text{.}$ It turns out to be true that in the limit this is about $1.1\text{.}$ Two examples don't establish a pattern, but if you do more examples you will find that the limit varies directly with the value of $a\text{:}$ bigger $a\text{,}$ bigger limit; smaller $a\text{,}$ smaller limit. As we can already see, some of these limits will be less than 1 and some larger than 1. Somewhere between $a=2$ and $a=3$ the limit will be exactly 1; the value at which this happens is called $e\text{,}$ so that

\begin{equation*} \lim_{\Delta x\to 0} {e^{\Delta x}-1\over \Delta x}=1\text{.} \end{equation*}

As you might guess from our two examples, $e$ is closer to 3 than to 2, and in fact $e\approx 2.718\text{.}$

Now we see that the function $\ds e^x$ has a truly remarkable property:

\begin{equation*} \begin{split} {d\over dx}e^x =\amp \lim_{\Delta x\to 0} {e^{x+\Delta x}-e^x\over \Delta x}\\ =\amp \lim_{\Delta x\to 0} {e^xe^{\Delta x}-e^x\over \Delta x}\\ =\amp \lim_{\Delta x\to 0} e^x{e^{\Delta x}-1\over \Delta x}\\ =\amp e^x\lim_{\Delta x\to 0} {e^{\Delta x}-1\over \Delta x}\\ =\amp e^x\end{split} \end{equation*}

That is, $\ds e^x$ is its own derivative, or in other words the slope of $\ds e^x$ is the same as its height, or the same as its second coordinate: The function $\ds f(x)=e^x$ goes through the point $\ds (z,e^z)$ and has slope $\ds e^z$ there, no matter what $z$ is. It is sometimes convenient to express the function $\ds e^x$ without an exponent, since complicated exponents can be hard to read. In such cases we use $\exp(x)\text{,}$ e.g., $\ds \exp(1+x^2)$ instead of $\ds e^{1+x^2}\text{.}$

What about the logarithm function? This too is hard, but as the cosine function was easier to do once the sine was done, so is the logarithm easier to do now that we know the derivative of the exponential function. Let's start with $\ds \log_e x\text{,}$ which as you probably know is often abbreviated $\ln x$ and called the “natural logarithm” function.

Consider the relationship between the two functions, namely, that they are inverses, that one “undoes” the other. Graphically this means that they have the same graph except that one is “flipped” or “reflected” through the line $y=x$ as shown in Figure 4.5.

This means that the slopes of these two functions are closely related as well: For example, the slope of $\ds e^x$ is $e$ at $x=1\text{;}$ at the corresponding point on the $\ln(x)$ curve, the slope must be $1/e\text{,}$ because the “rise” and the “run” have been interchanged. Since the slope of $\ds e^x$ is $e$ at the point $(1,e)\text{,}$ the slope of $\ln(x)$ is $1/e$ at the point $(e,1)\text{.}$

More generally, we know that the slope of $\ds e^x$ is $\ds e^z$ at the point $\ds (z,e^z)\text{,}$ so the slope of $\ln(x)$ is $\ds 1/e^z$ at $\ds (e^z,z)\text{,}$ as shown in Figure 4.6. In other words, the slope of $\ln x$ is the reciprocal of the first coordinate at any point; this means that the slope of $\ln x$ at $(x,\ln x)$ is $1/x\text{.}$ The upshot is:

\begin{equation*} {d\over dx}\ln x = {1\over x}\text{.} \end{equation*}

We have discussed this from the point of view of the graphs, which is easy to understand but is not normally considered a rigorous proof—it is too easy to be led astray by pictures that seem reasonable but that miss some hard point. It is possible to do this derivation without resorting to pictures, and indeed we will see an alternate approach soon.

Note that $\ln x$ is defined only for $x>0\text{.}$ It is sometimes useful to consider the function $\ln |x|\text{,}$ a function defined for $x\not=0\text{.}$ When $x\lt 0\text{,}$ $\ln |x|=\ln(-x)$ and

\begin{equation*} {d\over dx}\ln |x|={d\over dx}\ln (-x)={1\over -x}(-1)={1\over x}\text{.} \end{equation*}

Thus whether $x$ is positive or negative, the derivative is the same.

What about the functions $\ds a^x$ and $\ds \log_a x\text{?}$ We know that the derivative of $\ds a^x$ is some constant times $\ds a^x$ itself, but what constant? Remember that “the logarithm is the exponent” and you will see that $\ds a=e^{\ln a}\text{.}$ Then

\begin{equation*} a^x = (e^{\ln a})^x = e^{x\ln a}\text{,} \end{equation*}

and we can compute the derivative using the Chain Rule:

\begin{equation*} {d\over dx} a^x = {d\over dx}(e^{\ln a})^x = {d\over dx}e^{x\ln a} = (\ln a)e^{x\ln a} =(\ln a)a^x\text{.} \end{equation*}

The constant is simply $\ln a\text{.}$ Likewise we can compute the derivative of the logarithm function $\ds \log_a x\text{.}$ Since

\begin{equation*} x=e^{\ln x} \end{equation*}

we can take the logarithm base $a$ of both sides to get

\begin{equation*} \log_a(x)=\log_a(e^{\ln x})=\ln x \log_a e\text{.} \end{equation*}

Then

\begin{equation*} {d\over dx}\log_a x = {1\over x}\log_a e\text{.} \end{equation*}

This is a perfectly good answer, but we can improve it slightly. Since

\begin{equation*} \begin{split} a\amp =\amp e^{\ln a}\\ \log_a(a) \amp =\amp \log_a(e^{\ln a}) = \ln a\log_a e\\ 1\amp =\amp \ln a\log_a e\\ {1\over \ln a}\amp =\amp \log_a e\end{split}\text{,} \end{equation*}

we can replace $\ds \log_a e$ to get

\begin{equation*} {d\over dx}\log_a x = {1\over x\ln a}\text{.} \end{equation*}

You may if you wish memorize the formulas.

Because the “trick” $\ds a=e^{\ln a}$ is often useful, and sometimes essential, it may be better to remember the trick, not the formula.

###### Example4.60. Derivative of Exponential Function.

Compute the derivative of $\ds f(x)=2^x\text{.}$

Solution
\begin{equation*} \begin{split} {d\over dx}2^{x} =\amp {d\over dx}(e^{\ln 2})^x\\ =\amp {d\over dx}e^{x\ln 2}\\ =\amp \left({d\over dx} x\ln 2\right) e^{x\ln 2}\\ =\amp (\ln 2) e^{x\ln 2}=2^x\ln2\end{split} \end{equation*}
###### Example4.61. Derivative of Exponential Function.

Compute the derivative of $\ds f(x)=2^{x^2}=2^{(x^2)}\text{.}$

Solution
\begin{equation*} \begin{split} {d\over dx}2^{x^2} =\amp {d\over dx}e^{x^2\ln 2}\\ =\amp \left({d\over dx} x^2\ln 2\right) e^{x^2\ln 2}\\ =\amp (2\ln 2) x e^{x^2\ln 2}\\ = \amp (2\ln 2) x 2^{x^2}\end{split} \end{equation*}
###### Example4.62. Proving the General Power Rule.

Recall that we have not justified the Power Rule except when the exponent is a positive or negative integer. We can now show that

\begin{equation*} \ds{\frac{d}{dx}(x^n)=nx^{n-1}}\text{,} \end{equation*}

where $n$ is any real number.

Solution

We can use the exponential function to take care of other exponents.

\begin{equation*} \begin{split} {d\over dx}x^n =\amp {d\over dx}e^{n\ln x}\\ =\amp \left({d\over dx}n\ln x\right)e^{n\ln x}\\ =\amp \left(n{1\over x}\right)x^n \\ =\amp nx^{n-1}\end{split} \end{equation*}
###### Example4.63. Derivative of Exponential Function with Chain Rule.

Find the derivative of

\begin{equation*} g(t)=e^{2t^{2}+t} \end{equation*}
Solution

$g'(t) = e^{2t^{2}+t} \frac{d}{dt}(2t^{2}+t) = (4t+1)e^{2t^{2}+t}\text{.}$

###### Example4.64. Derivative of Exponential Function with Quotient Rule.

Find the derivative of

\begin{equation*} R(q) = \frac{q}{10^{q}+5} \end{equation*}
Solution

Using the Quotient Rule, we calculate

\begin{equation*} R'(q) = \frac{\frac{d}{dx}(q)(10^{q}+5) - q\frac{d}{dx}(10^{q}+5)}{(10^{q}+5)^{2}} \end{equation*}

Now, using the derivative formula for the exponential function with base $10\text{,}$ we find

\begin{equation*} \begin{split} R'(q) \amp = \frac{(1)(10^{q}+5) - q(\ln 10)(10^{q})}{(10^{q}+5)^{2}} \\ \amp = \frac{10^{q}+5 - (\ln 10)(q10^{q})}{(10^{q}+5)^{2}} \cdot \end{split} \end{equation*}
###### Example4.65. Derivative of Logarithmic Function with Chain Rule.

Find the derivative of

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

We use the Chain Rule and the formula for the derivative of a logarithmic function of base $2$ to find

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

Note that we must have $x > 5\text{.}$ Therefore, we need to be careful when we want to simplify the expression on the right side of the equation, and state the restriction as well:

\begin{equation*} y' = \frac{1}{2(\ln 2)(x-5)} \ \ \ x > 5\text{.} \end{equation*}
###### Example4.66. Derivative of Logarithmic Function with Chain Rule.

Find the derivative of

\begin{equation*} g(t)=\ln\left(t^{2}e^{-t^{2}}\right) \end{equation*}
Solution

To save a lot of work, we first simplify the given expression using properties of logarithms. We have

\begin{equation*} \begin{split} g(t) \amp = \ln\left(t^{2}e^{-t^{2}}\right)\\ \amp = \ln t^{2} + \ln e^{-t^{2}} \\ \amp = 2\ln t - t^{2}. \end{split} \end{equation*}

Therefore,

\begin{equation*} g'(t) = \frac{2}{t} - 2t = \frac{2(1-t^{2})}{t} \cdot \end{equation*}

### Subsection4.6.1The Fundamental Limit of Calculus

In Section 2.3 we defined $e$ to be the base so that the slope of the tangent line to the function $y=a^x$ at the point $x=0$ was equal to $1\text{,}$ i.e. $\frac{d(e^x)}{dx} \big\vert_{x=0} = 1\text{.}$ We will now see how we can write $e$ in terms of a limit.

Let $f(x) = \ln x\text{,}$ then $f'(x) = \frac{1}{x}$ and so $f'(1) = 1\text{.}$ We now write $f'(1)$ using the definition of derivative:

\begin{equation*} \begin{split} 1 \amp= f'(1) = \lim_{h\to 0} \frac{f(1+h)-f(1)}{h} = \lim\limits_{h\to 0}\frac{\ln(1+h)-\ln(1)}{h}\\ \amp= \lim\limits_{h\to 0} \frac{1}{h} \ln(1+h) = \lim\limits_{h\to 0} \ln(1+h)^{1/h}\end{split} \end{equation*}

Taking $e$ of both sides and using continuity of the exponential function gives:

\begin{equation*} e^1 = e^{\lim\limits_{h\to 0} \ln(1+h)^{1/h}} \iff e = \lim\limits_{h\to 0} e^{\ln (1+h)^{1/h}} = \lim\limits_{h\to 0} (1+h)^{1/h} \end{equation*}

For the second limit, we let $n=\frac{1}{h}\text{.}$ Then $h\to 0 \implies n \to \infty$ and $h = \frac{1}{n}\text{:}$

\begin{equation*} e = \lim\limits_{h\to 0} (1+h)^{1/h} = \lim\limits_{n\to \infty} \left(1+\frac{1}{n}\right)^n \end{equation*}
##### Exercises for Section 4.6.

Find the derivatives of the functions.

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

$\ds 2\ln(3)x3^{x^2}$
Solution

We rewrite $f(x)=3^{x^{2}}$ as $f(x)=e^{\ln 3^{x^{2}}}=e^{x^{2}\ln 3}\text{.}$ \begin{aligned}\diff{f}{x} \amp = e^{x^{2}\ln 3} \cdot \ln 3 \cdot 2x \\ \amp =2\ln(3)x3^{x^{2}}. \end{aligned}

2. $\ds f(x) = {\sin x \over e^x}$

$\ds {\cos x-\sin x \over e^x}$
Solution

Let $f(x) = \dfrac{\sin x}{e^{x}} = e^{-x}\sin x\text{.}$ Then its derivative is given by \begin{aligned}f'(x)\amp =e^{-x}\cos x - e^{-x}\sin x \amp = \frac{\cos x - \sin x}{e^{x}}\cdot \end{aligned}

3. $\ds f(t) = (e^t)^2$

$\ds 2e^{2t}$
Solution

\begin{aligned}f'(t) \amp = \diff{}{t} \left(e^t\right)^2 \\ \amp = 2e^t \diff{}{t} e^t \\ \amp = 2\left(e^{t}\right)^2 \\ \amp = 2 e^{2t} \end{aligned}

4. $\ds f(x) = \sin(e^x)$

$\ds e^x\cos(e^x)$
Solution

We have $f(x)=\sin\left(e^{x}\right)\text{.}$ We apply the chain rule, using the trigonometric and exponential derivative formulas to find $f'(x)=e^{x}\cos\left(e^{x}\right)\text{.}$

5. $\ds f(x) = e^{\sin x}$

$\ds \cos (x) e^{\sin x}$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} e^{\sin x} \\ \amp = e^{\sin x} \diff{}{x} \sin x \\ \amp = e^{\sin x} \cos x \end{aligned}

6. $\ds h(x) = x^{\sin x}$

$\ds x^{\sin x}\left(\cos x\ln x+{\sin x\over x}\right)$
Solution

We first rewrite $h(x) = x^{\sin x} = e^{\sin x \ln x}\text{.}$ \begin{aligned}h'(x) \amp = e^{\sin x \ln x} \diff{}{x} \sin x \ln x\\ \amp = e^{\sin x \ln x} \left(\ln x \cos x + \frac{\sin x}{x}\right) \end{aligned}

7. $\ds f(x) = x^3e^x$

$\ds 3x^2e^x+x^3e^x$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} x^3 e^x \\ \amp = e^x \diff{}{x} x^3 + x^3 \diff{}{x} e^x \\ \amp = e^x (3x^2) + x^3 e^x \\ \amp = e^x x^2 \left(3+x\right) \end{aligned}

8. $\ds g(t) = t+2^t$

$\ds 1+2^t\ln(2)$
Solution

\begin{aligned}g'(t) \amp = \diff{}{t} \left(t+2^t\right)\\ \amp = 1 + 2^t \ln 2 \end{aligned}

9. $\ds g(x) = (1/3)^{x^2}$

$\ds -2x\ln(3)(1/3)^{x^2}$
Solution

Let $g(x) = \left(1/3\right)^{x^2} = e^{x^2\ln(1/3)} = e^{-x^2 \ln(3) }\text{.}$ \begin{aligned}g'(x) \amp = \diff{}{x} e^{-x^2 \ln(3) } \\ \amp = e^{-x^2\ln (3)} \diff{}{x} \left(-x^2\ln (3)\right) \\ \amp = -2\ln(3) x e^{-x^2\ln(3)} \end{aligned}

10. $\ds f(s) = e^{4s}/s$

$\ds e^{4s}(4s-1)/s^2$
Solution

\begin{aligned}f'(s) \amp = \diff{}{s} \frac{e^{4s}}{s} \\ \amp = \frac{s\diff{}{s} e^{4s} - e^{4s} \diff{}{s} s}{s^2}\\ \amp = \frac{s (4e^{4s}) - e^{4s}}{s^2}\\ \amp = \frac{e^{4s} \left(4s-1\right)}{s^2} \end{aligned}

11. $\ds h(x) = \ln(x^3+3x)$

$\ds (3x^2+3)/(x^3+3x)$
Solution

\begin{aligned}\diff{h}{x} \amp =\diff{}{x}\left(\ln \left(x^{3}+3x\right)\right) \\ \amp = \frac{1}{x^{3}+3x}\cdot \diff{}{x}\left(x^{3}+3x\right) \\ \amp = \frac{3x^{2}+3}{x^{3}+3x} \end{aligned}

12. $\ds f(x) = \ln(\cos(x))$

$\ds -\tan(x)$
Solution

\begin{aligned}f'(x)\amp =\diff{}{x} \ln\left(\cos x\right)\\ \amp =\frac{1}{\cos x}(-\sin x)\\ \amp = -\tan x \end{aligned}

13. $\ds f(t) = \sqrt{\ln(t^2)}/t$

$\ds (1-\ln(t^2))/(t^2\sqrt{\ln(t^2)})$
Solution

Let's first apply the quotient rule. \begin{aligned}f'(t) \amp = \diff{}{t} \frac{\sqrt{\ln t^{2}}}{t} \amp = \frac{t\diff{}{t}\left(\sqrt{\ln t^{2}}\right) - \sqrt{\ln t^{2}}}{t^{2}}\cdot \end{aligned} Next, apply the chain rule. \begin{aligned}f'(t) \amp = \frac{t\cdot\frac{1}{2}(\ln t^{2})^{-1/2}\cdot\frac{2t}{t^{2}} - \sqrt{\ln t^{2}}}{t^{2}} \amp = \frac{1-\ln t^{2}}{t^{2}\sqrt{\ln t^{2}}}\cdot \end{aligned}

14. $\ds f(x) = \ln(\sec(x) + \tan(x))$

$\ds \frac{\tan x \sec x + \sec^2 x}{\sec x + \tan x}$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} \ln (\sec x + \tan x) \\ \amp = \frac{1}{\sec x + \tan x} \diff{}{x} \left(\sec x + \tan x\right) \\ \amp = \frac{\tan x \sec x + \sec^2 x}{\sec x + \tan x} \end{aligned}

15. $\ds g(x) = x^{\cos(x)}$

$\ds x^{\cos(x)}(\cos(x)/x-\cos(x)\ln(x))$
Solution

\begin{aligned}g'(x) \amp = \diff{}{x} x^{\cos x}\\ \amp = \diff{}{x} e^{\cos x \ln x} \\ \amp = e^{\cos x \ln x} \diff{}{x} \cos x \ln x \\ \amp = e^{\cos x \ln x} \left(\frac{\cos x }{x} - \sin x \ln x\right) \end{aligned}

16. $\ds h(s) = s\ln s$

$\ds s \frac{1}{s} + \ln s$
Solution

\begin{aligned}h'(s) \amp = \diff{}{s} s \ln s \\ \amp = s \frac{1}{s} + \ln s \end{aligned}

17. $\ds f(x) = \ln (\ln (3x) )$

$\ds \frac{3}{3x \ln(3x)}$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} \ln(\ln(3x)) \\ \amp = \frac{1}{\ln(3x)} \diff{}{x} \ln(3x)\\ \amp = \frac{1}{\ln (3x)} \cdot \frac{1}{3x} \cdot \diff{}{x} (3x) \\ \amp = \frac{1}{\ln (3x)} \cdot \frac{1}{3x} \cdot 3\\ \amp = \frac{3}{3x \ln(3x)} \end{aligned}

18. $\ds f(t) = {1+\ln (3t^2 )\over 1+ \ln(4t)}$

$\ds\frac{\left(1+\ln(3t^2)\right)\frac{1}{t} - \left(1+\ln(4t)\right) \frac{2}{t}}{\left(1+\ln(4t)\right)^2}$
Solution

\begin{aligned}f'(t) \amp = \diff{}{t} \left(\frac{1+\ln(3t^2)}{1+\ln(4t)}\right) \\ \amp = \frac{\left(1+\ln(3t^2)\right)\frac{4}{4t} - \left(1+\ln(4t)\right) \frac{6t}{3t^2}}{\left(1+\ln(4t)\right)^2}\\ \amp = \frac{\left(1+\ln(3t^2)\right)\frac{1}{t} - \left(1+\ln(4t)\right) \frac{2}{t}}{\left(1+\ln(4t)\right)^2} \end{aligned}

Find the second derivative of the function.

1. $\ds f(t)=3e^{-2t}-5e^{-t}$

$\ds f''(t)=12e^{-2t}-5e^{-t}$
Solution

\begin{aligned}f'(t) \amp = \diff{}{t} \left(3e^{-2t} - 5 e^{-t}\right) \\ \amp = 3 e^{-2t} \diff{}{t} (-2t) - 5e^{-t} \diff{}{t} (-t) \\ \amp = -6e^{-2t} +5e^{-t} f''(t) \amp = \diff{}{t} \left( -6e^{-2t} +5e^{-t}\right)\\ \amp = -6e^{-2t} \diff{}{t} (-2t) + 5e^{-t} \diff{}{t} (-t) \\ \amp = 12e^{-2t} - 5e^{-t} \end{aligned}

2. $\ds f(x)=x^{2}e^{-2x}$

$\ds f''(s)=2(2x^{2}-4x+1)e^{-2x}$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} x^2e^{-x} \\ \amp = e^{-x} \diff{}{x} x^2 + x^2 \diff{}{x} e^{-x} \\ \amp = 2xe^{-x} + x^2 e^{-x} \diff{}{x} (-x)\\ \amp = 2xe^{-x} - x^2 e^{-x} f''(x) \amp = \diff{}{x} \left(2xe^{-x} - x^2 e^{-x}\right)\\ \amp = \left(2e^{-x} -2xe^{-x}\right) - \left( f'(x)\right) \\ \amp = \left(2e^{-x} -2xe^{-x}\right) - \left( 2xe^{-x} - x^2 e^{-x}\right) \\ \amp = 2e^{-x} -4xe^{-x} - x^2 e^{-x} \end{aligned}

3. $\ds f(t)=(t-5)3^{t}$

$\ds f''(t)=\left(2\ln 3-5(\ln 3)^{2}\right)3^{t}+t(\ln 3)^{2}3^{t}$
Solution

\begin{aligned}f'(t) \amp = \diff{}{t} (t-5)3^t \\ \amp = 3^t \diff{}{t} (t-5) + (t-5) \diff{}{t} 3^t \\ \amp = 3^t + (t-5) \ln(3) 3^t f''(t) \amp = \diff{}{t} \left(3^t + (t-5) \ln(3) 3^t\right)\\ \amp = \ln(3) 3^t + \ln(3) \left(f'(t)\right) \\ \amp = \ln(3) 3^t + \ln(3) \left(3^t + (t-5) \ln(3) 3^t\right) \\ \amp = \ln(3) 3^t \left(2 +(t-5)\ln(3)\right) \end{aligned}

4. $\ds y=6^{\sqrt{x}}$

$\ds y''(x)=\dfrac{(\ln 6)^{2}\sqrt{x}6^{\sqrt{x}}-(\ln 6)6^{\sqrt{x}}}{4(\sqrt{x})^{3}}$
Solution

\begin{aligned}y'(x) \amp = \diff{}{x} 6^{\sqrt{x}}\\ \amp = \ln(6) 6^{\sqrt{x}} \diff{}{x} \sqrt{x} \\ \amp = \ln(6) 6^{\sqrt{x}} \frac{1}{2\sqrt{x}} \end{aligned}

Find an equation of the tangent line to the graph of $y=e^{2x-3}$ at the point $\left(\frac{3}{2},1\right)\text{.}$

$y=2x-2$
Solution

Let $f(x) = e^{2x-3}\text{.}$ We first differentiate:

\begin{equation*} f'(x) = e^{2x-3} \diff{}{x} (2x-3) = 2e^{2x-3}\text{.} \end{equation*}

Therefore, at the point $(3/2,1)\text{,}$ the slope of the tangent line is

\begin{equation*} f'(3/2) = 2e^{3-3} = 2\text{.} \end{equation*}

Hence, an equation of the tangent line is

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

Find an equation of the tangent line to the graph of $y=e^{-x^{2}}$ at the point $\left(1,\frac{1}{e}\right)\text{.}$

$y=-\frac{2x}{e}+\frac{3}{e}$
Solution

Let $f(x) = e^{-x^2}\text{.}$ Then its derivative is

\begin{equation*} f'(x) = e^{-x^2} \diff{}{x} (-x^2) = -2x e^{-x^2}\text{.} \end{equation*}

So at the point of tangency $\left(1, \frac{1}{e}\right)\text{,}$

\begin{equation*} f'(1) = -2 e^{-1} = \frac{-2}{e}\text{.} \end{equation*}

Hence, an equation of the tangent line is

\begin{equation*} y - \frac{1}{e} = \frac{-2}{e} (x-1)\text{.} \end{equation*}

Find the value of $a$ so that the tangent line to $y=\ln(x)$ at $x=a$ is a line through the origin. Sketch the resulting situation.

$e$
Solution

The slope of the tangent line to $y=\ln(x)$ at the point $x=a$ is

\begin{equation*} \diff{}{x}\ln(x)\big\vert_{x=a} = \frac{1}{a}\text{.} \end{equation*}

Further, the tangent line passes through the point $(a, \ln(a))\text{.}$ This gives the equation of the tangent line as a function of $a\text{:}$

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

So if we want this tangent line to pass through the point $(0,0)\text{,}$ we must have

\begin{equation*} -\ln(a) = \frac{1}{a}(-a) \implies \ln(a)=1 \implies a = e\text{.} \end{equation*}

We verify with a sketch:

If $\ds f(x) = \ln(x^3 + 2)$ compute $\ds f'(e^{1/3})\text{.}$

$f'(e^{1/3}) = \frac{3e^{2/3}}{e+2}$
Solution

We are given that $f(x) = \ln(x^3+2)\text{.}$ Therefore, by the Chain Rule:

\begin{equation*} f'(x) = \frac{1}{x^3+2} \diff{}{x} (x^3+2) = \frac{3x^2}{x^3+2}\text{.} \end{equation*}

And so

\begin{equation*} f'(e^{1/3}) = \frac{3e^{2/3}}{e+2}\text{.} \end{equation*}

The unit selling price $p$ (in dollars) and the quantity $x$ demanded (in pairs) of a certain brand of women's gloves is given by the demand equation

\begin{equation*} p=100e^{-0.0001x} \ \ \ 0\leq x \leq 20,000 \end{equation*}
1. Find the revenue function $R$ (Hint: $R(x) = px$).

$R(x) = px = 100xe^{-0.0001x}\text{,}$ for $0 \leq x \leq 200,000\text{.}$
2. Find the marginal revenue function $R'\text{.}$

$R'(x) = 100\diff{}{x} xe^{-0.0001x} = 100\left(e^{-0.0001x} + x(-0.0001)e^{-0.0001x}\right) = 100e^{-0.0001x}(1-0.0001x)\text{.}$
3. What is the marginal revenue when $x=10\text{?}$

$R'(10) = 100e^{-0.00x}(1-0.001)\text{.}$ Thus, when 10 pairs of gloves are demanded, the marginal revenue is approximately $99.80. The monthly demand for a certain brand of table wine is given by the demand equation \begin{equation*} p=240\left(1-\frac{3}{3+e^{-0.0005x}}\right) \end{equation*} where $p$ denotes the wholesale price per case (in dollars) and $x$ denotes the number of cases demanded. 1. Find the rate of change of the price per case when $x=1000\text{.}$ Answer -$$0.0168$ per case
Solution

We first differentiate:

\begin{equation*} \begin{split} p'(x) \amp = \diff{}{x} 240 \left(1-\frac{3}{3+e^{-0.0005x}}\right) \\ \amp = -240 \diff{}{x} \left(\frac{3}{3+e^{-0.0005x}}\right) \\ \amp = -720 \diff{}{x} \left(3+e^{-0.0005x}\right)^{-1} \\ \amp = 720 \left(3+e^{-0.0005x}\right)^{-2} \diff{}{x}\left(3+e^{-0.0005x}\right)\\ \amp = -720 \left(3+e^{-0.0005x}\right)^{-2} \left(-0.0005 e^{-0.0005x}\right) \\ \amp = -\frac{9 e^{-0.0005x}}{25 \left(3+e^{-0.0005x}\right)^{2} } \end{split} \end{equation*}

And so when 1000 cases are demanded, we have

\begin{equation*} p'(1000) = -\frac{9 e^{-.5}}{25 \left(3+e^{-.5x}\right)^{2} } \approx -0.0168\text{,} \end{equation*}

which means that the unit price is decreasing by approximately %0.0168 per case.

2. What is the price per case when $x=1000\text{?}$

$$40.36$ per case. Solution The unit price when 1000 cases are demanded is given by \begin{equation*} p(1000) = 240 \left(1-\frac{3}{3+e^{-.5}}\right) \approx 40.36\text{,} \end{equation*} or$40.36.

The price of a certain commodity in dollars per unit time (measured in weeks) is given by

\begin{equation*} p=8+4e^{-2t}+te^{-2t} \end{equation*}
1. What is the price of the commodity at $t=0\text{?}$

$$12$/unit Solution At $t=0\text{,}$ the unit price of the commodity is $p=8+4e^{0}+(0)e^{0} = 8+4(1) = 12\text{,}$ or$12.

2. How fast is the price of the commodity changing at $t=0\text{?}$

-$$7$/week Solution $\diff{p}{t} = -8e^{-2t}+e^{-2t}-2te^{-2t}\text{.}$ At $t=0\text{,}$ $\diff{p}{t}=-7\text{.}$ Therefore, the price of the commodity is decreasing by$7 per week at this time.

3. Find the equilibrium price of the commodity (Hint: It's given by $\lim\limits_{t\to\infty}p\text{.}$ Also use the fact that $\lim\limits_{t\to\infty} te^{-2t} = 0\text{.}$

\$$8$/unit
Solution

$\lim\limits_{t\to\infty}p = 8 + 4\lim\limits_{t\to\infty} e^{-2t} + \lim\limits_{t\to\infty}te^{-2t}\text{.}$ We use the fact that $e^{-2t} \to 0$ as $t \to \infty$ and the hint to see that $\lim\limits_{t\to\infty}p = 8\text{.}$

The percent of households using online banking may be approximated by the formula

\begin{equation*} f(t)=1.5e^{0.78t} \ \ \ 0 \leq t \leq 4 \end{equation*}

where $t$ is measured in years, with $t=0$ corresponding to the beginning of $2000\text{.}$

1. What is the projected percent of households using online banking at the beginning of $2003\text{?}$

$14.23$%
Solution

Since $t$ is measured in years since 2000, the beginning of the year 2003 corresponds to $t=3\text{.}$ Therefore, since

\begin{equation*} f(3) = 1.5e^{0.75(3)} \approx 14.23\text{,} \end{equation*}

approximately 14.23% of all households were projected to be using online banking by the beginning of 2003.

2. How fast will the projected percent of households using online banking be changing at the beginning of $2003\text{?}$

$10.67$%/yr
Solution

We differentiate:

\begin{equation*} f'(t) = 1.5e^{0.75t} (0.75) = 1.125 e^{0.75t}\text{.} \end{equation*}

Therefore,

\begin{equation*} f'(3) = 1.125e^{0.75(3)} \approx 10.67\text{.} \end{equation*}

Hence, it was projected that the rate of households using online banking by the beginning of 2003 was increasing at a rate of about 10.67% per year.

3. How fast will the rate of the projected percent of households using online banking be changing at the beginning of $2003\text{?}$ (Hint: we want $f''(3)\text{.}$ Why?)

$8$%/yr/yr
Solution

We differentiate again:

\begin{equation*} f''(t) = 1.125 e^{0.75t} (0.75) = 0.84375 e^{0.75t}\text{.} \end{equation*}

And so

\begin{equation*} f''(3) \approx 8.006\text{,} \end{equation*}

which means that the rate at which the rate of households using online banking was increasing was projected to be about 8% per year.

The average energy consumption of the typical refrigerator/freezer manufactured by York Industries is approximately

\begin{equation*} C(t)=1486e^{-0.073t}+500 \ \ \ 0 \leq t \leq 20 \end{equation*}

killowatt-hours (kWh) per year, where $t$ is measured in years, with $t=0$ corresponding to $1972\text{.}$

1. What was the average energy consumption of the York refrigerator/freezer at the beginning of $1972\text{?}$

$1986$ kWh/yt
Solution

$C(0) = 1486e^{0}+500 = 1986\text{.}$ Therefore, the average energy consumption at the beginning of 1972 was 1986 kWh/yr.

2. What is the rate of change of the average energy consumption?

$C'(t)=-108.48e^{-0.073t}$
Solution

The rate of change of the average energy comsumption at the beginning of year $t$ is

\begin{equation*} C'(t) = -0.073(1486e^{-0.073t}) = -108.478 e^{-0.073t}\text{.} \end{equation*}
3. All refrigerator/freezers manufactured as of Janurary $1\text{,}$ $1990\text{,}$ must meet a $950$-kWh/yr maximum energy-consumption standard. Show that the York refigerator/freezer satisfies this requirement.

$C(18)=899.35 \lt 950$
for all $t\text{,}$ this means that the average energy consumption per year is decreasing from the years 1990 to 1992. Therefore, all fridges manufactured as of Janurary 1990 meet the standard.