# Calculus/Differentiation/Basics of Differentiation/Exercises
## Find the Derivative by Definition

Find the derivative of the following functions using the limit definition of the derivative.
$f(x)=x^{2}$
$2x$
$2x$
$f(x)=2x+2$
$2$
$2$
$f(x)={\frac {x^{2}}{2}}$
$x$
$x$
$f(x)=2x^{2}+4x+4$
$4x+4$
$4x+4$
$f(x)={\sqrt {x+2}}$
${\frac {1}{2{\sqrt {x+2}}}}$
${\frac {1}{2{\sqrt {x+2}}}}$
$f(x)={\frac {1}{x}}$
$-{\frac {1}{x^{2}}}$
$-{\frac {1}{x^{2}}}$
$f(x)={\frac {3}{x+1}}$
${\frac {-3}{(x+1)^{2}}}$
${\frac {-3}{(x+1)^{2}}}$
$f(x)={\frac {1}{\sqrt {x+1}}}$
${\frac {-1}{2(x+1)^{3/2}}}$
${\frac {-1}{2(x+1)^{3/2}}}$
$f(x)={\frac {x}{x+2}}$
${\frac {2}{(x+2)^{2}}}$
${\frac {2}{(x+2)^{2}}}$
Solutions

## Prove the Constant Rule

$c$
${\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]$
${\begin{aligned}{\frac {d}{dx}}[c\cdot f(x)]&=\lim _{\Delta x\to 0}{\frac {c\cdot f(x+\Delta x)-c\cdot f(x)}{\Delta x}}\\&=c\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=c{\frac {d}{dx}}[f(x)]\end{aligned}}$
${\begin{aligned}{\frac {d}{dx}}[c\cdot f(x)]&=\lim _{\Delta x\to 0}{\frac {c\cdot f(x+\Delta x)-c\cdot f(x)}{\Delta x}}\\&=c\lim _{\Delta x\to 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}\\&=c{\frac {d}{dx}}[f(x)]\end{aligned}}$
Solutions

## Find the Derivative by Rules

Find the derivative of the following functions:

### Power Rule

$f(x)=2x^{2}+4$
$f'(x)=4x$
$f'(x)=4x$
$f(x)=3{\sqrt[{3}]{x}}\,$
$f'(x)={\frac {1}{\sqrt[{3}]{x^{2}}}}$
$f'(x)={\frac {1}{\sqrt[{3}]{x^{2}}}}$
$f(x)=2x^{5}+8x^{2}+x-78$
$f'(x)=10x^{4}+16x+1$
$f'(x)=10x^{4}+16x+1$
$f(x)=7x^{7}+8x^{5}+x^{3}+x^{2}-x$
$f'(x)=49x^{6}+40x^{4}+3x^{2}+2x-1$
$f'(x)=49x^{6}+40x^{4}+3x^{2}+2x-1$
$f(x)={\frac {1}{x^{2}}}+3x^{\frac {1}{3}}$
$f'(x)={\frac {-2}{x^{3}}}+{\frac {1}{\sqrt[{3}]{x^{2}}}}$
$f'(x)={\frac {-2}{x^{3}}}+{\frac {1}{\sqrt[{3}]{x^{2}}}}$
$f(x)=3x^{15}+{\frac {x^{2}}{17}}+{\frac {2}{\sqrt {x}}}$
$f'(x)=45x^{14}+{\frac {2x}{17}}-{\frac {1}{x{\sqrt {x}}}}$
$f'(x)=45x^{14}+{\frac {2x}{17}}-{\frac {1}{x{\sqrt {x}}}}$
$f(x)={\frac {3}{x^{4}}}-{\sqrt[{4}]{x}}+x$
$f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[{4}]{x^{3}}}}}+1$
$f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[{4}]{x^{3}}}}}+1$
$f(x)=6x^{1/3}-x^{0.4}+{\frac {9}{x^{2}}}$
$f'(x)={\frac {2}{\sqrt[{3}]{x^{2}}}}-{\frac {0.4}{x^{0.6}}}-{\frac {18}{x^{3}}}$
$f'(x)={\frac {2}{\sqrt[{3}]{x^{2}}}}-{\frac {0.4}{x^{0.6}}}-{\frac {18}{x^{3}}}$
$f(x)={\frac {1}{\sqrt[{3}]{x}}}+{\sqrt {x}}$
$f'(x)={\frac {-1}{3{\sqrt[{3}]{x^{4}}}}}+{\frac {1}{2{\sqrt {x}}}}$
$f'(x)={\frac {-1}{3{\sqrt[{3}]{x^{4}}}}}+{\frac {1}{2{\sqrt {x}}}}$
Solutions

### Product Rule

$f(x)=(x^{4}+4x+2)(2x+3)$
$10x^{4}+12x^{3}+16x+16$
$10x^{4}+12x^{3}+16x+16$
$f(x)=(2x-1)(3x^{2}+2)$
$18x^{2}-6x+4$
$18x^{2}-6x+4$
$f(x)=(x^{3}-12x)(3x^{2}+2x)$
$15x^{4}+8x^{3}-108x^{2}-48x$
$15x^{4}+8x^{3}-108x^{2}-48x$
$f(x)=(2x^{5}-x)(3x+1)$
$36x^{5}+10x^{4}-6x-1$
$36x^{5}+10x^{4}-6x-1$
$f(x)=(5x^{2}+3)(2x+7)$
$30x^{2}+70x+6$
$30x^{2}+70x+6$
$f(x)=3x^{2}(5x^{2}+1)^{4}$
$6x(25x^{2}+1)(5x^{2}+1)^{3}$
$6x(25x^{2}+1)(5x^{2}+1)^{3}$
$f(x)=x^{3}(2x^{2}-x+4)^{4}$
$x^{2}(2x^{2}-x+4)^{3}(22x^{2}-7x+12)$
$x^{2}(2x^{2}-x+4)^{3}(22x^{2}-7x+12)$
$f(x)=5x^{2}(x^{3}-x+1)^{3}$
$5x(x^{3}-x+1)^{2}(11x^{3}-5x+2)$
$5x(x^{3}-x+1)^{2}(11x^{3}-5x+2)$
$f(x)=(2-x)^{6}(5+2x)^{4}$
$2(x-2)^{5}(2x+5)^{3}(10x+7)$
$2(x-2)^{5}(2x+5)^{3}(10x+7)$
Solutions

### Quotient Rule

$f(x)={\frac {2x+1}{x+5}}$
$f'(x)={\frac {9}{(x+5)^{2}}}$
$f'(x)={\frac {9}{(x+5)^{2}}}$
$f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}$
$f'(x)={\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}$
$f'(x)={\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}$
$f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}$
$f'(x)={\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}$
$f'(x)={\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}$
$d(u)={\frac {u^{3}+2}{u^{3}}}$
$d'(u)=-{\frac {6}{u^{4}}}$
$d'(u)=-{\frac {6}{u^{4}}}$
$f(x)={\frac {x^{2}+x}{2x-1}}$
$f'(x)={\frac {2x^{2}-2x-1}{(2x-1)^{2}}}$
$f'(x)={\frac {2x^{2}-2x-1}{(2x-1)^{2}}}$
$f(x)={\frac {x+1}{2x^{2}+2x+3}}$
$f'(x)={\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}$
$f'(x)={\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}$
$f(x)={\frac {16x^{4}+2x^{2}}{x}}$
$f'(x)=48x^{2}+2$
$f'(x)=48x^{2}+2$
$f(x)={\frac {8x^{3}+2}{5x+5}}$
$f'(x)={\frac {2(8x^{3}+12x^{2}-1)}{5(x+1)^{2}}}$
$f'(x)={\frac {2(8x^{3}+12x^{2}-1)}{5(x+1)^{2}}}$
$f(x)={\frac {(3x-2)^{2}}{\sqrt {x}}}$
$f'(x)={\frac {(3x-2)(9x+2)}{2x{\sqrt {x}}}}$
$f'(x)={\frac {(3x-2)(9x+2)}{2x{\sqrt {x}}}}$
$f(x)={\frac {\sqrt {x}}{2x-1}}$
$f'(x)={\frac {-(2x+1)}{2{\sqrt {x}}(2x-1)^{2}}}$
$f'(x)={\frac {-(2x+1)}{2{\sqrt {x}}(2x-1)^{2}}}$
$f(x)={\frac {4x-3}{x+2}}$
$f'(x)={\frac {11}{(x+2)^{2}}}$
$f'(x)={\frac {11}{(x+2)^{2}}}$
$f(x)={\frac {4x+3}{2x-1}}$
$f'(x)={\frac {-10}{(2x-1)^{2}}}$
$f'(x)={\frac {-10}{(2x-1)^{2}}}$
$f(x)={\frac {x^{2}}{x+3}}$
$f'(x)={\frac {x(x+6)}{(x+3)^{2}}}$
$f'(x)={\frac {x(x+6)}{(x+3)^{2}}}$
$f(x)={\frac {x^{5}}{3-x}}$
$f'(x)={\frac {x^{4}(-4x+15)}{(3-x)^{2}}}$
$f'(x)={\frac {x^{4}(-4x+15)}{(3-x)^{2}}}$
Solutions

### Chain Rule

$f(x)=(x+5)^{2}$
$f'(x)=2(x+5)$
$f'(x)=2(x+5)$
$g(x)=(x^{3}-2x+5)^{2}$
$g'(x)=2(x^{3}-2x+5)(3x^{2}-2)$
$g'(x)=2(x^{3}-2x+5)(3x^{2}-2)$
$f(x)={\sqrt {1-x^{2}}}$
$f'(x)=-{\frac {x}{\sqrt {1-x^{2}}}}$
$f'(x)=-{\frac {x}{\sqrt {1-x^{2}}}}$
$f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}$
$f'(x)={\frac {6(4x^{3}+1)(2x+4)^{2}-(2x+4)^{3}(12x^{2})}{(4x^{3}+1)^{2}}}$
$f'(x)={\frac {6(4x^{3}+1)(2x+4)^{2}-(2x+4)^{3}(12x^{2})}{(4x^{3}+1)^{2}}}$
$f(x)=(2x+1){\sqrt {2x+2}}$
$f'(x)=2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}$
$f'(x)=2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}$
$f(x)={\frac {2x+1}{\sqrt {2x+2}}}$
$f'(x)={\frac {2x+3}{(2x+2){\sqrt {2x+2}}}}$
$f'(x)={\frac {2x+3}{(2x+2){\sqrt {2x+2}}}}$
$f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}$
$f'(x)={\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)$
$f'(x)={\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)$
$f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}$
$f'(x)={\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}$
$f'(x)={\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}$
$f(x)={\sqrt {x^{3}+1}}(x^{2}-1)$
$f'(x)={\frac {3x^{2}(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}$
$f'(x)={\frac {3x^{2}(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}$
$f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}$
$f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)$
$f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)$
$f(x)={\sqrt {1+x^{2}}}$
$f'(x)={\frac {x}{\sqrt {1+x^{2}}}}$
$f'(x)={\frac {x}{\sqrt {1+x^{2}}}}$
Solutions

### Exponentials

$f(x)=(3x^{2}+e)e^{2x}$
$f'(x)=6xe^{2x}+2e^{2x}(3x^{2}+e)$
$f'(x)=6xe^{2x}+2e^{2x}(3x^{2}+e)$
$f(x)=e^{2x^{2}+3x}$
$f'(x)=(4x+3)e^{2x^{2}+3x}$
$f'(x)=(4x+3)e^{2x^{2}+3x}$
$f(x)=e^{e^{2x^{2}+1}}$
$f'(x)=4xe^{2x^{2}+1+e^{2x^{2}+1}}$
$f'(x)=4xe^{2x^{2}+1+e^{2x^{2}+1}}$
$f(x)=4^{x}$
$f'(x)=\ln(4)4^{x}$
$f'(x)=\ln(4)4^{x}$
Solutions

### Logarithms

$f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln(x)$
$f'(x)=3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-3}}{2{\sqrt {x^{3}-2}}}}+{\frac {1}{x}}$
$f'(x)=3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-3}}{2{\sqrt {x^{3}-2}}}}+{\frac {1}{x}}$
$f(x)=\ln(x)-2e^{x}+{\sqrt {x}}$
$f'(x)={\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}$
$f'(x)={\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}$
$f(x)=\ln(\ln(x^{3}(x+1)))$
$f'(x)={\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}$
$f'(x)={\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}$
$f(x)=\ln(2x^{2}+3x)$
$f'(x)={\frac {4x+3}{2x^{2}+3x}}$
$f'(x)={\frac {4x+3}{2x^{2}+3x}}$
$f(x)=\log _{4}(x)+2\ln(x)$
$f'(x)={\frac {1}{x\ln(4)}}+{\frac {2}{x}}$
$f'(x)={\frac {1}{x\ln(4)}}+{\frac {2}{x}}$
Solutions

### Trigonometric functions

$f(x)=3e^{x}-4\cos(x)-{\frac {\ln(x)}{4}}$
$f'(x)=3e^{x}+4\sin(x)-{\frac {1}{4x}}$
$f'(x)=3e^{x}+4\sin(x)-{\frac {1}{4x}}$
$f(x)=\sin(x)+\cos(x)$
$f'(x)=\cos(x)-\sin(x)$
$f'(x)=\cos(x)-\sin(x)$
Solutions

## More Differentiation

${\frac {d}{dx}}[(x^{3}+5)^{10}]$
$30x^{2}(x^{3}+5)^{9}$
$30x^{2}(x^{3}+5)^{9}$
${\frac {d}{dx}}[x^{3}+3x]$
$3x^{2}+3$
$3x^{2}+3$
${\frac {d}{dx}}[(x+4)(x+2)(x-3)]$
$(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)$
$(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)$
${\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]$
$-{\frac {x+2}{3x^{3}}}$
$-{\frac {x+2}{3x^{3}}}$
${\frac {d}{dx}}[3x^{3}]$
$9x^{2}$
$9x^{2}$
${\frac {d}{dx}}[x^{4}\sin(x)]$
$4x^{3}\sin(x)+x^{4}\cos(x)$
$4x^{3}\sin(x)+x^{4}\cos(x)$
${\frac {d}{dx}}[2^{x}]$
$\ln(2)2^{x}$
$\ln(2)2^{x}$
${\frac {d}{dx}}[e^{x^{2}}]$
$2xe^{x^{2}}$
$2xe^{x^{2}}$
${\frac {d}{dx}}[e^{2^{x}}]$
$\ln(2)2^{x}e^{2^{x}}$
$\ln(2)2^{x}e^{2^{x}}$
Solutions

## Implicit Differentiation

Use implicit differentiation to find y'
$x^{3}+y^{3}=xy$
$y'={\frac {y-3x^{2}}{3y^{2}-x}}$
$y'={\frac {y-3x^{2}}{3y^{2}-x}}$
$(2x+y)^{4}+3x^{2}+3y^{2}={\frac {x}{y}}+1$
$y'={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}$
$y'={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}$
Solutions

## Logarithmic Differentiation

Use logarithmic differentiation to find ${\frac {dy}{dx}}$:
$y=x({\sqrt[{4}]{1-x^{3}}})$
$y'={\sqrt[{4}]{1-x^{3}}}+{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}$
$y'={\sqrt[{4}]{1-x^{3}}}+{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}$
$y={\sqrt {x+1 \over 1-x}}\,$
$y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}\left({\frac {1}{x+1}}+{\frac {1}{1-x}}\right)$
$y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}\left({\frac {1}{x+1}}+{\frac {1}{1-x}}\right)$
$y=(2x)^{2x}$
$y'=(2x)^{2x}(2\ln(2x)+2)$
$y'=(2x)^{2x}(2\ln(2x)+2)$
$y=(x^{3}+4x)^{3x+1}$
$y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})$
$y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})$
$y=(6x)^{\cos(x)+1}$
$y'=6x^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})$
$y'=6x^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})$
Solutions

## Equation of Tangent Line

For each function, $f$ , (a) determine for what values of $x$ the tangent line to $f$ is horizontal and (b) find an equation of the tangent line to $f$ at the given point.
$f(x)={\frac {x^{3}}{3}}+x^{2}+5,\qquad (3,23)$
$x=0,-2$
$y=15x-22$
$x=0,-2$
$y=15x-22$
$f(x)=x^{3}-3x+1,\qquad (1,-1)$
$x=\pm 1$
$y=-1$
$x=\pm 1$
$y=-1$
$f(x)={\frac {2x^{3}}{3}}+x^{2}-12x+6,\qquad (0,6)$
$x=2,-3$
$y=-12x+6$
$x=2,-3$
$y=-12x+6$
$f(x)=2x+{\frac {1}{\sqrt {x}}},\qquad (1,3)$
$x=2^{-{\frac {4}{3}}}$
$y={\frac {3x}{2}}+{\frac {3}{2}}$
$x=2^{-{\frac {4}{3}}}$
$y={\frac {3x}{2}}+{\frac {3}{2}}$
$f(x)=(x^{2}+1)(2-x),\qquad (2,0)$
$x=1,{\frac {1}{3}}$
$y=-5x+10$
$x=1,{\frac {1}{3}}$
$y=-5x+10$
$f(x)={\frac {2x^{3}}{3}}+{\frac {5x^{2}}{2}}+2x+1,\qquad \left(3,{\frac {95}{2}}\right)$
$x=-{\frac {1}{2}},-2$
$y=35x-{\frac {115}{2}}$
$x=-{\frac {1}{2}},-2$
$y=35x-{\frac {115}{2}}$
$(x-y-1)^{3}=x$
$y={\frac {2x}{3}}-{\frac {5}{3}}$
$y={\frac {2x}{3}}-{\frac {5}{3}}$
$e^{xy}+x^{2}=y^{2}$
$y=-2x+2$
$y=-2x+2$
Solutions

## Higher Order Derivatives

$3x^{4}+3x^{2}+2x$
$36x^{2}+6$
$36x^{2}+6$
base case: Consider the zeroth-order polynomial, $c$ . ${\frac {dc}{dx}}=0$ induction step: Suppose that the n-th derivative of a (n-1)th order polynomial is 0. Consider the n-th order polynomial, $f(x)$ . We can write $f(x)=cx^{n}+P(x)$ where $P(x)$ is a (n-1)th polynomial.
${\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0$
base case: Consider the zeroth-order polynomial, $c$ . ${\frac {dc}{dx}}=0$ induction step: Suppose that the n-th derivative of a (n-1)th order polynomial is 0. Consider the n-th order polynomial, $f(x)$ . We can write $f(x)=cx^{n}+P(x)$ where $P(x)$ is a (n-1)th polynomial.
${\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0$
Solutions

## Advanced Understanding of Derivatives

$f^{\prime }(x)$
$f(x)$
$-f(x)$
$-f^{\prime }(x)$
Suppose $f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}$. Let $g(x)=-f(x)$. ${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$
$f^{\prime }(x)$
$f(x)$
$-f^{\prime }(x)$
$-f(x)$
$\blacksquare$
Suppose $f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}$. Let $g(x)=-f(x)$. ${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$
$f^{\prime }(x)$
$f(x)$
$-f^{\prime }(x)$
$-f(x)$
$\blacksquare$
$f(x)$
$-2\leq x\leq 2$
$f(-2)=f(2)<0$
$f(x)$
    (a) the Intermediate Value Theorem;
    (b) Rolle's Theorem;
    (c) the Extreme Value Theorem.
These are examples only. More valid solutions may exist.
    (a) $f(x)$ is continuous. Ergo, the intermediate value theorem applies. There exists some $f(d)=c\in \left[f(-2),0\right]$ such that $f(-2)<c<0$, where $d\in \left[-2,2\right]$.
    (b) Rolle's Theorem does not apply for a non-differentiable function.
    (c) $f(x)$ is continuous. Ergo, the extreme value theorem applies. There exists a $-2\leq c\leq 2$ so that $f(c)>f(x)$ for all $x\in \left[-2,2\right]$.
These are examples only. More valid solutions may exist.
    (a) $f(x)$ is continuous. Ergo, the intermediate value theorem applies. There exists some $f(d)=c\in \left[f(-2),0\right]$ such that $f(-2)<c<0$, where $d\in \left[-2,2\right]$.
    (b) Rolle's Theorem does not apply for a non-differentiable function.
    (c) $f(x)$ is continuous. Ergo, the extreme value theorem applies. There exists a $-2\leq c\leq 2$ so that $f(c)>f(x)$ for all $x\in \left[-2,2\right]$.
$g(x)=f^{-1}(x)$
$f^{-1}(x)$
$f(x)$
$f(x)$
$g^{\prime }(x)$
$g^{\prime }(x)$
$g^{\prime }(x)={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}$
$g^{\prime }(x)={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}$
$f(x)={\begin{cases}2x^{2}+4,&x<-1\\ax^{3}-2x,&x\geq -1\end{cases}}$
$a$
Find a value, if possible, for $a$ that allows each of the following to be true. If not possible, prove that it cannot be done.
    (a) The function $f(x)$ is continuous but non-differentiable.
    (b) The function $f(x)$ is both continuous and differentiable.
    (a) $a=-4$.
    (b) There is no $a\in \mathbb {R}$ that allows the following to be true. Proof in solutions.
    (a) $a=-4$.
    (b) There is no $a\in \mathbb {R}$ that allows the following to be true. Proof in solutions.
Solutions
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