Find the derivatives of the following functions from definition.

1. $f(x)=(2x+3)^2$ Answer

$4(2x+3)$
Solution

We use the definition of the derivative:

\begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h) -f(x)}{h}\\ \amp = \lim_{h\to 0} \frac{(2(x+h)+3)^2 - (2x+3)^2}{h}\\ \amp = \lim_{h\to 0} \frac{(4(x+h)^2 + 9 + 12(x+h)) - (4x^2+9+12x)}{h}\\ \amp = \lim_{h\to 0} \frac{8xh +4h^2 + 12h}{h} \\ \amp = \lim_{h\to 0} (8x+4h+12)\\ \amp = 8x + 12 \end{split} \end{equation*}

2. $g(x)=x^{3/2}$

$\frac{3}{2}x^{1/2}$
Solution

We use the definition of the derivative:

\begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h) -f(x)}{h}\\ \amp = \lim_{h\to 0} \frac{(x+h)^{3/2} - x^{3/2}}{h}\\ \amp = \lim_{h\to 0} \frac{(x+h) \sqrt{h + x} - x\sqrt{x}}{h}\\ \amp = \lim_{h\to 0} \frac{x\sqrt{h+x} + h \sqrt{h+x} - x\sqrt{x}}{h}\\ \amp = \lim_{h\to 0} \frac{h\sqrt{h+x} + x(\sqrt{h+x}-\sqrt{x})}{h}\\ \amp = \lim_{h\to 0} \frac{h\sqrt{h+x}}{h} + \lim_{h\to 0} \frac{x(\sqrt{h+x}-\sqrt{x})}{h} \\ \amp = \lim_{h\to 0} \sqrt{h+x} + \lim_{h\to 0} \frac{x(x+h-x)}{h(\sqrt{x+h}+\sqrt{x})} \\ \amp = \sqrt{x} + \lim_{h\to 0} \frac{xh}{h(\sqrt{x+h}+\sqrt{x})} \\ \amp = \sqrt{x} + \lim_{h\to 0} \frac{x}{\sqrt{x+h}+\sqrt{x}} \\ \amp = \sqrt{x} + \frac{x}{2\sqrt{x}}\\ \amp =\frac{3\sqrt{x}}{2} \end{split} \end{equation*}

Let $f(x)=\left\{ \begin{array}{cc} x^{3} \amp \text{ if } x\leq 1 \\ 5x-x^{2} \amp \text{ if } x>1 \end{array} \right. \text{.}$ Use the definition of the derivative to find $f^{\prime}(1)\text{.}$

\begin{equation*} f'(x) = \begin{cases}3x^2 \amp x \lt 1 \\ 5-2x \amp x > 1 \end{cases} \end{equation*}
Solution

We use the definition of the derivative. Let $x \lt 1\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h) -f(x)}{h}\\ \amp = \lim_{h\to 0} \frac{(x+h)^3 - x^3}{h}\\ \amp = \lim_{h\to 0} \frac{(h^3 + 3 h^2 x + 3 h x^2 + x^3) - x^3}{h}\\ \amp = \lim_{h\to 0} \frac{h(h^2+3hx+3x^2)}{h} \\ \amp = \lim_{h\to 0} (h^2+3hx+3x^2) \\ \amp = 3x^2 \end{split} \end{equation*}

Now let $x > 1\text{:}$

\begin{equation*} \begin{split} f'(x) \amp = \lim_{h\to 0} \frac{f(x+h) -f(x)}{h}\\ \amp = \lim_{h\to 0} \frac{(5(x+h)-(x+h)^2) - (5x-x^2)}{h}\\ \amp = \lim_{h\to 0} \frac{5x + 5h -x^2-2xh-h^2 - 5x + x^2}{h}\\ \amp = \lim_{h\to 0} \frac{h(5-2x-h)}{h} \\ \amp = \lim_{h\to 0} (5-2x-h)\\ \amp = 5-2x \end{split} \end{equation*}

The function is not differentiable at $x=1\text{.}$ Therefore,

\begin{equation*} f'(x) = \begin{cases}3x^2 \amp x \lt 1 \\ 5-2x \amp x > 1 \end{cases} \end{equation*}

Differentiate the following functions.

1. $y=7x^4-7\pi^4+\dfrac{1}{\pi\sqrt[3]{x}}$

$28x^3-\dfrac{1}{3\pi x^{4/3}}$
Solution

\begin{aligned}y' \amp = \diff{}{x} \left(7x^4-7\pi^4+\frac{1}{\pi^3\sqrt{x}}\right)\\ \amp = 7(4)x^3-(0)-\frac{1}{2}\frac{1}{\pi^3x^{3/2}} \\ \amp = 28x^3-\frac{1}{2\pi^3x^{3/2}} \end{aligned}

2. $f(x)=\dfrac{1-\sqrt{x}}{1+\sqrt{x}}$

$-\dfrac{1}{\sqrt{x}(1+\sqrt{x})^2}$
Solution

\begin{aligned}\diff{}{x}f(x) \amp = \diff{}{x} \left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)\\ \amp = \frac{(1+\sqrt{x})\diff{}{x}\left(1-\sqrt{x}\right)-(1-\sqrt{x})\diff{}{x}\left(1+\sqrt{x}\right)}{(1+\sqrt{x})^2}\\ \amp =\frac{(1+\sqrt{x})\left(-\frac{1}{2\sqrt{x}}\right)-\left(1-\sqrt{x}\right)\left(\frac{1}{2\sqrt{x}}\right)}{(1+\sqrt{x})^2}\\ \amp = \frac{-(1+\sqrt{x})-(1-\sqrt{x})}{2\sqrt{x}(1+\sqrt{x})^2}\\ \amp =-\frac{1}{\sqrt{x}(1+\sqrt{x})^2} \end{aligned}

3. $f(x)=\vert x-1\vert+\vert x+2\vert$

\begin{equation*} f'(x) = \begin{cases}2 \amp x \lt 1 \\ 0 \amp -2 \lt x \lt 1 \\ -2 \amp x > -2 \\ \text{ DNE } \amp x=-1,2 \end{cases} \end{equation*}
Solution

We first write $f(x)$ as a piecewise-defined function:

\begin{equation*} f(x) = \begin{cases}(x-1)+(x+2) = 2x+1 \amp x \geq 1 \\ -(x-1)+x+2 = 3 \amp -2 \lt x \lt 1 \\ -(x-1)-(x+2) = -2x-1 \amp x \geq -2 \end{cases} \end{equation*}

Therefore, we can differentiate each branch of $f(x)$ noting that the derivative does not exist at the corners:

\begin{equation*} f'(x) = \begin{cases}2 \amp x \lt 1 \\ 0 \amp -2 \lt x \lt 1 \\ -2 \amp x > -2 \\ \text{ DNE } \amp x=-1,2 \end{cases} \end{equation*}
4. $f(x)=x^2\sin x\cos x$

$2x\sin x\cos x+x^2\cos^2 x-x^2\sin^2 x$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} x^2\sin x\cos x\\ \amp = 2x\sin x \cos x + x^2(\cos x)\cos x + x^2\sin x (-\sin x) \\ \amp = 2x\sin x \cos x + x^2\cos^2x - x^2\sin^2 x \end{aligned}

5. $y=\dfrac{x\sin x}{1+\sin x}$

$\dfrac{(\sin x+x\cos x)(1+\sin x)-x\sin x\cos x}{(1+\sin x)^2}$
Solution

\begin{aligned}y' \amp = \diff{}{x} \frac{x\sin x}{1+\sin x} \\ \amp = \frac{(\sin x + x\cos x)(1+\sin x) - (\cos x)(x\sin x)}{(1+\sin x)^2} \end{aligned}

6. $g(x)=\sqrt{2+\dfrac{3}{\sqrt{x}}}$

$-\dfrac{3}{4x^{3/2}}\left(2+\dfrac{3}{\sqrt{x}}\right)^{-1/2}$
Solution

\begin{aligned}g'(x) \amp = \diff{}{x} \sqrt{2+\frac{3}{\sqrt{x}}} \\ \amp = \frac{1}{2\sqrt{2+\frac{3}{\sqrt{x}}}} \diff{}{x} \left(2+\frac{3}{\sqrt{x}}\right) \\ \amp = \frac{1}{2\sqrt{2+\frac{3}{\sqrt{x}}}} \cdot \frac{3}{2x^{3/2}} \\ \amp = \frac{3}{4x^{3/2}\sqrt{2+\frac{3}{\sqrt{x}}}} \end{aligned}

7. $y=\sqrt[3]{x^4+x^2+1}+\dfrac{1}{(x^3-x+4)^5}$

$\frac{1}{3}(x^4+x^2+1)^{-2/3}(4x^3+2x)-\dfrac{5(3x^2-1)}{(x^3-x+4)^6}$
Solution

\begin{aligned}y' \amp = \diff{}{x} \left(\sqrt[3]{x^4+x^2+1}+\frac{1}{(x^3-x+4)^5}\right)\\ \amp = \frac{1}{3} (x^4+x^2+1)^{-2/3}\diff{}{x} (x^4+x^2+1) -5(x^3-x+4)^{-6}\diff{}{x}(x^3-x+4) \\ \amp = \frac{4x^3+2x}{3}(x^4+x^2+1)^{-2/3} - 5(x^3-x+4)^{-6} (3x^2-1) \end{aligned}

8. $y=\sin^3 x-\sin(x^3)$

$3\sin^2 x\cos x-3x^2\cos(x^3)$
Solution

\begin{aligned}y' \amp = \diff{}{x} \left(\sin^3 x - \sin(x^3)\right)\\ \amp = 3\sin^2 x \diff{}{x} \sin x - \cos(x^3) \diff{}{x} x^3 \\ \amp = 3\sin^2 x \cos x - 3x^2\cos(x^3) \end{aligned}

9. $F(x)=\sec^4 x+\tan^4 x$

$4\sec^4 x\tan x + 4\tan^3 x\sec^2 x$
Solution

\begin{aligned}F'(x) \amp = \diff{}{x}\left(\sec^4x+\tan^4x\right) \\ \amp = 4\sec^3x(\tan x \sec x) + 4\tan^3 x(\sec^2 x)\\ \amp =4\sec^4 x \tan x+ 4\tan^3\sec^2 x \end{aligned}

10. $y=\cos^2\left(\dfrac{1-x}{1+x}\right)$

$\dfrac{4}{(1+x)^2}\cos\left(\dfrac{1-x}{1+x}\right)\sin\left(\dfrac{1-x}{1+x}\right)$
Solution

\begin{aligned}y'\amp =\diff{}{x} \cos^2\left(\frac{1-x}{1+x}\right)\\ \amp = -2\cos\left(\frac{1-x}{1+x}\right)\sin\left(\frac{1-x}{1+x}\right)\cdot \frac{-(1+x)-(1-x)}{(1+x)^2}\\ \amp = \frac{4}{(1+x)^2}\cos\left(\frac{1-x}{1+x}\right)\sin\left(\frac{1-x}{1+x}\right) \end{aligned}

11. $y=\tan(\sin(x^2+\sec^2 x))$

$(2x+2\sec^2 x\tan x)\sec^2(\sin(x^2+\sec^2 x))\cos(x^2+\sec^2 x)$
Solution

\begin{aligned}y' \amp = \diff{}{x} \tan(\sin(x^2+\sec^2 x)) \\ \amp = \sec^2(\sin(x^2+\sec^2 x)) \diff{}{x} \sin(x^2+\sec^2 x))\\ \amp = \sec^2(\sin(x^2+\sec^2 x)) \cos(x^2+\sec^2 x)) \diff{}{x} (x^2+\sec^2 x)\\ \amp = \sec^2(\sin(x^2+\sec^2 x)) \cos(x^2+\sec^2 x)) (2x + 2\sec x \tan x \sec x) \\ \amp = \sec^2(\sin(x^2+\sec^2 x)) \cos(x^2+\sec^2 x)) (2x + 2\sec^2 x \tan x) \end{aligned}

12. $y=\dfrac{1}{2+\sin\frac{\pi}{x}}$

$-\dfrac{\pi\cos\frac{\pi}{x}}{x^2(2+\sin\frac{\pi}{x})^2}$

Differentiate the following functions.

1. $y=e^{3x}+e^{-x}+e^2$

$3e^{3x}-e^{-x}$
Solution

$y' = \diff{}{x} \left(e^{3x}+e^{-x}+e^2\right) = 3e^{3x}-e^x\text{.}$

2. $y=e^{2x}\cos 3x$

$2e^{2x}\cos 3x-3e^{2x}\sin 3x$
Solution

$y' = \diff{}{x} e^{2x}\cos(3x) = 2e^{2x}\cos(3x) - 3e^{2x}\sin(x)$

3. $f(x)=\tan(x+e^x)$

$(1+e^x)\sec^2(x+e^x)$
Solution

$y' = \diff{}{x} \tan(x+e^x) = (1+e^x)\sec^2(x+e^x)$

4. $g(x)=\dfrac{e^x}{e^x+2}$

$2e^x/(e^x+2)^2$
Solution

$g'(x) = \diff{}{x} \dfrac{e^x}{e^x+2} = \dfrac{e^x(e^x+2)-e^{2x}}{(e^x+2)^2} = \dfrac{2e^x}{(e^x+2)^2}$

5. $y=\ln(2+\sin x)-\sin(2+\ln x)$

$\dfrac{\cos x}{2+\sin x}-\dfrac{\cos(2+\ln x)}{x}$
Solution

$y' = \diff{}{x} \ln(2+\sin x) - \sin(2+\ln x) = \frac{\cos x}{2+\sin(x)} - \dfrac{\cos(2+\ln x)}{x}$

6. $f(x)=e^{x^{\pi}}+x^{\pi^{e}}+\pi^{e^x}$

$e^{x^{\pi}}\cdot\pi x^{\pi-1}+\pi^e x^{\pi^e-1}+\pi^{e^x}\ln\pi\cdot e^x$
Solution

$f'(x) = \diff{}{x} \left(e^{x^{\pi}} + x^{\pi^e} + \pi^{e^x}\right) =e^{x^{\pi}}\cdot\pi x^{\pi-1}+\pi^e x^{\pi^e-1}+\pi^{e^x}\ln\pi\cdot e^x$

7. $y=\log _{a}(b^x)+b^{\log_a x}\text{,}$ where $a$ and $b$ are positive real numbers and $a\neq 1\text{.}$ Answer
$\log_{a}b+(\log_{a}b) x^{(\log_{a}b)-1}$
Solution

$y' = \diff{}{x} \log_a (b^x) + b^{\log_a x} = \log_{a}b+(\log_{a}b) x^{(\log_{a}b)-1}$

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

$(x^2+1)^{x^3+1}\left(3x^2\ln(x^2+1)+\dfrac{2x(x^3+1)}{x^2+1}\right)$
Solution

$y' = \diff{}{x} (x^2+1)^{x^3+1} = (x^2+1)^{x^3+1}\left(3x^2\ln(x^2+1)+\dfrac{2x(x^3+1)}{x^2+1}\right)$

9. $y=(x^2+e^x)^{1/\ln x}$

$\dfrac{(x^2+e^x)^{1/\ln x}}{(\ln x)^2}\left(\dfrac{2x+e^x}{x^2+e^x}\ln x-\dfrac{ln(x^2+e^x)}{x}\right)$
Solution

\begin{aligned}y \amp = (x^2+e^x)^{1/\ln x}\\ \ln(y) \amp = \frac{1}{\ln(x)} \ln(x^2+e^x)\\ \diff{}{x} \ln(y) \amp = \diff{}{x} \frac{1}{\ln(x)} \ln(x^2+e^x) \\ \frac{y'}{y} \amp = \frac{1}{(\ln(x))^2} \left(\ln(x) \frac{2x+e^x}{x^2+e^x} - \frac{\ln(x^2+e^x)}{x} \right)\\ y' \amp = \frac{y}{(\ln(x))^2} \left(\ln(x) \frac{2x+e^x}{x^2+e^x} - \frac{\ln(x^2+e^x)}{x} \right)\\ y' \amp = \frac{(x^2+e^x)^{1/\ln(x)}}{(\ln(x))^2} \left( \frac{\ln(x) (2x+e^x)}{x^2+e^x} - \frac{\ln(x^2+e^x)}{x}\right) \end{aligned}

10. $y=\dfrac{x\sqrt{x^2+x+1}}{(2+\sin x)^4(3x+5)^7}$

$\dfrac{x\sqrt{x^2+x+1}}{(2+\sin x)^4 (3x+5)^7} \left(\dfrac{1}{x}+\dfrac{2x+1}{2(x^2+x+1)}-\dfrac{4\cos x}{2+\sin x}-\dfrac{21}{3x+5}\right)$
Solution

\begin{aligned}\ln(y) \amp = \ln \left(\frac{x(x^2+x+1)^{1/2}}{(2+\sin(x))^4(3x+5)^7}\right) \\ \amp = \ln\left(x (x^2+x+1)^{1/2}\right)-\ln\left((2+\sin(x))^4(3x+5)^7\right) \\ \amp =\left[\ln(x) + \frac{1}{2}\ln(x^2+x+1)\right] - \left[ 4\ln(2+\sin(x))+7\ln(3x+5)\right] \end{aligned} We now differentiate: \begin{aligned}\frac{y'}{y} \amp =\diff{}{x} \left[\ln(x) + \frac{1}{2}\ln(x^2+x+1)\right] - \left[ 4\ln(2+\sin(x))+7\ln(3x+5)\right] \\ \amp = y\left[\frac{1}{x} + \frac{2x+1}{2(x^2+x+1)} - \frac{4\cos(x)}{2+\sin(x)} - \frac{21}{3x+5} \right]\\ \amp =\frac{x\sqrt{x^2+x+1}}{(2+\sin(x))^4(3x+5)^7} \left[ \frac{1}{x} + \frac{2x+1}{2(x^2+x+1)} - \frac{4\cos(x)}{2+\sin(x)} - \frac{21}{3x+5} \right] \end{aligned}

Find $\dfrac{dy}{dx}$ if $y$ is a differentiable function that satisfy the given equation.

1. $x^2+xy+y^2=7$

$-(2x+y)/(x+2y)$
Solution

\begin{aligned}\amp \diff{}{x} \left(x^2+xy+y^2\right) = \diff{}{x} (7)\\ \amp 2x + y + xy'+2yy' = 0 \\ \amp (x+2y)y' = -2x-y \\ \amp y' = -\frac{2x+y}{x+2y} \end{aligned}

2. $x^2+y^2=(2x^2+2y^2-x)^2$

$\dfrac{x-(2x^2+2y^2-x) (4x-1)}{4y(2x^2+2y^2-x)-y}$
Solution

\begin{aligned}\amp \diff{}{x} \left(x^2+y^2\right) = \diff{}{x} \left(2x^2+2y^2-x\right)^2 \\ \amp 2x + 2yy' = 2\left(2x^2+2y^2-x\right) \left(4x + 4yy' - 1\right) \\ \amp yy' -4yy'\left(2x^2+2y^2-x\right) = -x +\left(2x^2+2y^2-x\right) (4x-1) \\ \amp y' = \frac{-x +\left(2x^2+2y^2-x\right) (4x-1)} { y-4y\left(2x^2+2y^2-x\right) } \end{aligned}

3. $x^2\sin y+y^3=\cos x$

$-\dfrac{\sin x+2x\sin y}{x^2\cos y+3y^2}$
Solution

\begin{aligned}\amp \diff{}{x} (x^2\sin y + y^3) = \diff{}{x} \cos x\\ \amp 2x \sin y + x^2\cos y y' + 3y^2 y' = -\sin x \\ \amp x^2\cos y y' + 3y^2 y' = -\sin x - 2x \sin y\\ \amp y' = \frac{-\sin x - 2x \sin y}{x^2\cos y + 3y^2} \end{aligned}

4. $x^2+xe^y=2y+e^x$

$\dfrac{2x+e^y-e^x}{2-xe^y}$
Solution

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

Differentiate the following functions.

1. $y=x\sin^{-1}x$

$\sin^{-1}x+x/\sqrt{1-x^2}$
Solution

\begin{aligned}\diff{y}{x} \amp = \diff{}{x} x\sin^{-1} x\\ \amp = \sin^{-1} x + \frac{x}{\sqrt{1-x^2}} \end{aligned}

2. $f(x)=\dfrac{\sin^{-1}x}{\cos^{-1}x}$

$\dfrac{\cos^{-1}x+\sin^{-1}x}{(\cos^{-1}x)^2 \sqrt{1-x^2}}$
Solution

\begin{aligned}f'(x) \amp = \diff{}{x} \frac{\sin^{-1} x}{\cos^{-1} x} \\ \amp = \frac{\cos^{-1} x\diff{}{x} \sin^{-1} x -\sin^{-1} x \diff{}{x} \cos^{-1} x}{(\cos^{-1} x)^2}\\ \amp = \frac{\frac{\cos^{-1} x}{\sqrt{1-x^2}} +\frac{\sin^{-1} x}{\sqrt{1-x^2}} }{(\cos^{-1} x)^2}\\ \amp = \frac{\cos^{-1} x + \sin^{-1} x}{(\cos^{-1} x)^2\sqrt{1-x^2}} \end{aligned}

3. $g(x)=\tan^{-1}\left(\dfrac{x}{a}\right)\text{,}$ where $a>0$

$a/(x^2+a^2)$
\begin{aligned}g'(x) \amp = \diff{}{x} \tan^{-1}\left(\frac{x}{a}\right) \\ \amp = \frac{1}{1+x^2/a^2} \diff{}{x} \frac{x}{a} \\ \amp = \frac{1}{a+x^2/a}\\ \amp = \frac{a}{x^2+a^2} \end{aligned}
4. $y=x\tan^{-1}x-\dfrac{1}{2}\ln(x^2+1)$
$\tan^{-1}x$
\begin{aligned}y' \amp = \diff{}{x} x\tan^{-1} x - \frac{1}{2} \ln(x^2+1) \\ \amp = \tan^{-1} x + \frac{x}{1+x^2} - \frac{2x}{2(x^2+1)}\\ \amp = \tan^{-1} x \end{aligned}