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

$f(x)=x^{2}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{2}-x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{2}+2x\Delta x+\Delta x^{2}-x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}2x+\Delta x\\&=\mathbf {2x} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{2}-x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{2}+2x\Delta x+\Delta x^{2}-x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}2x+\Delta x\\&=\mathbf {2x} \end{aligned}}$
$f(x)=2x+2\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)+2]-(2x+2)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+2-2x-2}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\mathbf {2} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)+2]-(2x+2)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+2-2x-2}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\mathbf {2} \end{aligned}}$
$f(x)={\frac {1}{2}}x^{2}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{2}}(x+\Delta x)^{2}-{\frac {1}{2}}x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{2}}(x^{2}+2x\Delta x+\Delta x^{2})-{\frac {1}{2}}x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\frac {x^{2}}{2}}+{\frac {2x\Delta x}{2}}+{\frac {\Delta x^{2}}{2}}-{\frac {x^{2}}{2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{2\Delta x}}\\&=\lim _{\Delta x\to 0}x+{\frac {\Delta x}{2}}\\&=\mathbf {x} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{2}}(x+\Delta x)^{2}-{\frac {1}{2}}x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{2}}(x^{2}+2x\Delta x+\Delta x^{2})-{\frac {1}{2}}x^{2}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {{\frac {x^{2}}{2}}+{\frac {2x\Delta x}{2}}+{\frac {\Delta x^{2}}{2}}-{\frac {x^{2}}{2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x\Delta x+\Delta x^{2}}{2\Delta x}}\\&=\lim _{\Delta x\to 0}x+{\frac {\Delta x}{2}}\\&=\mathbf {x} \end{aligned}}$
$f(x)=2x^{2}+4x+4\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)^{2}+4(x+\Delta x)+4]-(2x^{2}+4x+4)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {4x\Delta x+2\Delta x^{2}+4\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}4x+2\Delta x+4\\&=\mathbf {4x+4} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {[2(x+\Delta x)^{2}+4(x+\Delta x)+4]-(2x^{2}+4x+4)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {4x\Delta x+2\Delta x^{2}+4\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}4x+2\Delta x+4\\&=\mathbf {4x+4} \end{aligned}}$
$f(x)={\sqrt {x+2}}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\sqrt {x+\Delta x+2}}-{\sqrt {x+2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}({\frac {{\sqrt {x+\Delta x+2}}-{\sqrt {x+2}}}{\Delta x}})({\frac {{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}{{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}})\\&=\lim _{\Delta x\to 0}{\frac {x+\Delta x+2-x-2}{\Delta x({\sqrt {x+\Delta x+2}}+{\sqrt {x+2}})}}\\&=\lim _{\Delta x\to 0}{\frac {\Delta x}{\Delta x({\sqrt {x+\Delta x+2}}+{\sqrt {x+2}})}}\\&=\lim _{\Delta x\to 0}{\frac {1}{{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}}\\&=\mathbf {\frac {1}{2{\sqrt {x+2}}}} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\sqrt {x+\Delta x+2}}-{\sqrt {x+2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}({\frac {{\sqrt {x+\Delta x+2}}-{\sqrt {x+2}}}{\Delta x}})({\frac {{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}{{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}})\\&=\lim _{\Delta x\to 0}{\frac {x+\Delta x+2-x-2}{\Delta x({\sqrt {x+\Delta x+2}}+{\sqrt {x+2}})}}\\&=\lim _{\Delta x\to 0}{\frac {\Delta x}{\Delta x({\sqrt {x+\Delta x+2}}+{\sqrt {x+2}})}}\\&=\lim _{\Delta x\to 0}{\frac {1}{{\sqrt {x+\Delta x+2}}+{\sqrt {x+2}}}}\\&=\mathbf {\frac {1}{2{\sqrt {x+2}}}} \end{aligned}}$
$f(x)={\frac {1}{x}}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{x+\Delta x}}-{\frac {1}{x}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {x-x-\Delta x}{x(x+\Delta x)}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {-\Delta x}{x\Delta x(x+\Delta x)}}\\&=\lim _{\Delta x\to 0}{\frac {-1}{x(x+\Delta x)}}\\&=\mathbf {-{\frac {1}{x^{2}}}} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{x+\Delta x}}-{\frac {1}{x}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {x-x-\Delta x}{x(x+\Delta x)}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {-\Delta x}{x\Delta x(x+\Delta x)}}\\&=\lim _{\Delta x\to 0}{\frac {-1}{x(x+\Delta x)}}\\&=\mathbf {-{\frac {1}{x^{2}}}} \end{aligned}}$
$f(x)={\frac {3}{x+1}}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {3}{x+\Delta x+1}}-{\frac {3}{x+1}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {3x+3-(3x+3\Delta x+3)}{(x+1)(x+\Delta x+1)}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {-3\Delta x}{\Delta x(x+1)(x+\Delta x+1)}}\\&=\lim _{\Delta x\to 0}{\frac {-3}{(x+1)(x+\Delta x+1)}}\\&=\mathbf {\frac {-3}{(x+1)^{2}}} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {3}{x+\Delta x+1}}-{\frac {3}{x+1}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {3x+3-(3x+3\Delta x+3)}{(x+1)(x+\Delta x+1)}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {-3\Delta x}{\Delta x(x+1)(x+\Delta x+1)}}\\&=\lim _{\Delta x\to 0}{\frac {-3}{(x+1)(x+\Delta x+1)}}\\&=\mathbf {\frac {-3}{(x+1)^{2}}} \end{aligned}}$
$f(x)={\frac {1}{\sqrt {x+1}}}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{\sqrt {x+\Delta x+1}}}-{\frac {1}{\sqrt {x+1}}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {{\sqrt {x+1}}-{\sqrt {x+\Delta x+1}}}{{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}({\frac {{\sqrt {x+1}}-{\sqrt {x+\Delta x+1}}}{\Delta x{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}}})({\frac {{\sqrt {x+1}}+{\sqrt {x+\Delta x+1}}}{{\sqrt {x+1}}+{\sqrt {x+\Delta x+1}}}})\\&=\lim _{\Delta x\to 0}{\frac {x+1-(x+\Delta x+1)}{\Delta x{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}({\sqrt {x+\Delta x+1}}+{\sqrt {x+1}})}}\\&=\lim _{\Delta x\to 0}{\frac {-1}{{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}({\sqrt {x+\Delta x+1}}+{\sqrt {x+1}})}}\\&={\frac {-1}{(x+1)(2{\sqrt {x+1}})}}\\&=\mathbf {\frac {-1}{2(x+1)^{3/2}}} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {1}{\sqrt {x+\Delta x+1}}}-{\frac {1}{\sqrt {x+1}}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\frac {{\sqrt {x+1}}-{\sqrt {x+\Delta x+1}}}{{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}({\frac {{\sqrt {x+1}}-{\sqrt {x+\Delta x+1}}}{\Delta x{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}}})({\frac {{\sqrt {x+1}}+{\sqrt {x+\Delta x+1}}}{{\sqrt {x+1}}+{\sqrt {x+\Delta x+1}}}})\\&=\lim _{\Delta x\to 0}{\frac {x+1-(x+\Delta x+1)}{\Delta x{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}({\sqrt {x+\Delta x+1}}+{\sqrt {x+1}})}}\\&=\lim _{\Delta x\to 0}{\frac {-1}{{\sqrt {x+\Delta x+1}}{\sqrt {x+1}}({\sqrt {x+\Delta x+1}}+{\sqrt {x+1}})}}\\&={\frac {-1}{(x+1)(2{\sqrt {x+1}})}}\\&=\mathbf {\frac {-1}{2(x+1)^{3/2}}} \end{aligned}}$
$f(x)={\frac {x}{x+2}}\,$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {x+\Delta x}{x+\Delta x+2}}-{\frac {x}{x+2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)(x+2)-x(x+\Delta x+2)}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {x^{2}+2x+x\Delta x+2\Delta x-x^{2}-x\Delta x-2x}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {2}{(x+\Delta x+2)(x+2)}}\\&=\mathbf {\frac {2}{(x+2)^{2}}} \end{aligned}}$
${\begin{aligned}f'(x)&=\lim _{\Delta x\to 0}{\frac {{\frac {x+\Delta x}{x+\Delta x+2}}-{\frac {x}{x+2}}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)(x+2)-x(x+\Delta x+2)}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {x^{2}+2x+x\Delta x+2\Delta x-x^{2}-x\Delta x-2x}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x(x+\Delta x+2)(x+2)}}\\&=\lim _{\Delta x\to 0}{\frac {2}{(x+\Delta x+2)(x+2)}}\\&=\mathbf {\frac {2}{(x+2)^{2}}} \end{aligned}}$

## Prove the Constant Rule

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

## Find the Derivative by Rules
### Power Rule

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

### Product Rule

$f(x)=(x^{4}+4x+2)(2x+3)\,$
$f'(x)=(4x^{3}+4)(2x+3)+(x^{4}+4x+2)(2)=\mathbf {10x^{4}+12x^{3}+16x+16}$
$f'(x)=(4x^{3}+4)(2x+3)+(x^{4}+4x+2)(2)=\mathbf {10x^{4}+12x^{3}+16x+16}$
$f(x)=(2x-1)(3x^{2}+2)\,$
$f'(x)=(2)(3x^{2}+2)+(2x-1)(6x)=\mathbf {18x^{2}-6x+4}$
$f'(x)=(2)(3x^{2}+2)+(2x-1)(6x)=\mathbf {18x^{2}-6x+4}$
$f(x)=(x^{3}-12x)(3x^{2}+2x)\,$
$f'(x)=(3x^{2}-12)(3x^{2}+2x)+(x^{3}-12x)(6x+2)=\mathbf {15x^{4}+8x^{3}-108x^{2}-48x}$
$f'(x)=(3x^{2}-12)(3x^{2}+2x)+(x^{3}-12x)(6x+2)=\mathbf {15x^{4}+8x^{3}-108x^{2}-48x}$
$f(x)=(2x^{5}-x)(3x+1)\,$
$f'(x)=(10x^{4}-1)(3x+1)+(2x^{5}-x)(3)=\mathbf {36x^{5}+10x^{4}-6x-1}$
$f'(x)=(10x^{4}-1)(3x+1)+(2x^{5}-x)(3)=\mathbf {36x^{5}+10x^{4}-6x-1}$

### Quotient Rule

$f(x)={\frac {2x+1}{x+5}}\,$
$f'(x)={\frac {(x+5)(2)-(2x+1)}{(x+5)^{2}}}=\mathbf {\frac {9}{(x+5)^{2}}}$
$f'(x)={\frac {(x+5)(2)-(2x+1)}{(x+5)^{2}}}=\mathbf {\frac {9}{(x+5)^{2}}}$
$f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}\,$
$f'(x)={\frac {(3x^{2}+1)(12x^{3}+2)-(3x^{4}+2x+2)(6x)}{(3x^{2}+1)^{2}}}=\mathbf {\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}$
$f'(x)={\frac {(3x^{2}+1)(12x^{3}+2)-(3x^{4}+2x+2)(6x)}{(3x^{2}+1)^{2}}}=\mathbf {\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}$
$f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}\,$
$f'(x)={\frac {(x+2)({\frac {3}{2}}{\sqrt {x}})-(x^{\frac {3}{2}}+1)}{(x+2)^{2}}}=\mathbf {\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}$
$f'(x)={\frac {(x+2)({\frac {3}{2}}{\sqrt {x}})-(x^{\frac {3}{2}}+1)}{(x+2)^{2}}}=\mathbf {\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}$
$d(u)={\frac {u^{3}+2}{u^{3}}}\,$
$d'(u)={\frac {u^{3}(3u^{2})-(u^{3}+2)(3u^{2})}{u^{6}}}=\mathbf {-{\frac {6}{u^{4}}}}$
$d'(u)={\frac {u^{3}(3u^{2})-(u^{3}+2)(3u^{2})}{u^{6}}}=\mathbf {-{\frac {6}{u^{4}}}}$
$f(x)={\frac {x^{2}+x}{2x-1}}\,$
$f'(x)={\frac {(2x-1)(2x+1)-(x^{2}+x)(2)}{(2x-1)^{2}}}=\mathbf {\frac {2x^{2}-2x-1}{(2x-1)^{2}}}$
$f'(x)={\frac {(2x-1)(2x+1)-(x^{2}+x)(2)}{(2x-1)^{2}}}=\mathbf {\frac {2x^{2}-2x-1}{(2x-1)^{2}}}$
$f(x)={\frac {x+1}{2x^{2}+2x+3}}\,$
$f'(x)={\frac {(2x^{2}+2x+3)-(x+1)(4x+2)}{(2x^{2}+2x+3)^{2}}}=\mathbf {\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}$
$f'(x)={\frac {(2x^{2}+2x+3)-(x+1)(4x+2)}{(2x^{2}+2x+3)^{2}}}=\mathbf {\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}$
$f(x)={\frac {16x^{4}+2x^{2}}{x}}\,$
$f'(x)={\frac {x(64x^{3}+4x)-(16x^{4}+2x^{2})}{x^{2}}}=\mathbf {48x^{2}+2}$
$f'(x)={\frac {x(64x^{3}+4x)-(16x^{4}+2x^{2})}{x^{2}}}=\mathbf {48x^{2}+2}$

### Chain Rule

$f(x)=(x+5)^{2}\,$
$f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)}$
$f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)}$
$g(x)=(x^{3}-2x+5)^{2}\,$
$g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5$
$g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)}$
$g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5$
$g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)}$
$f(x)={\sqrt {1-x^{2}}}\,$
$f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}}$
$f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}}$
$f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}\,$
$f(x)={\frac {N(x)}{D(x)}};\quad N(x)=g(h(x));\quad g(x)=x^{3};\quad h(x)=2x+4;\quad D(x)=4x^{3}+1$
$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$ $N'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=3(2x+4)^{2}(2)=6(2x+4)^{2}$ $D'(x)=12x^{2}$
$f'(x)={\frac {(4x^{3}+1)6(2x+4)^{2}-(2x+4)^{3}12x^{2}}{(4x^{3}+1)^{2}}}=\mathbf {\frac {6(4x^{3}+1)(2x+4)^{2}-12x^{2}(2x+4)^{3}}{(4x^{3}+1)^{2}}}$
$f(x)={\frac {N(x)}{D(x)}};\quad N(x)=g(h(x));\quad g(x)=x^{3};\quad h(x)=2x+4;\quad D(x)=4x^{3}+1$
$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$ $N'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=3(2x+4)^{2}(2)=6(2x+4)^{2}$ $D'(x)=12x^{2}$
$f'(x)={\frac {(4x^{3}+1)6(2x+4)^{2}-(2x+4)^{3}12x^{2}}{(4x^{3}+1)^{2}}}=\mathbf {\frac {6(4x^{3}+1)(2x+4)^{2}-12x^{2}(2x+4)^{3}}{(4x^{3}+1)^{2}}}$
$f(x)=(2x+1){\sqrt {2x+2}}\,$
$f(x)=A(x)B(x);\quad A(x)=2x+1;\quad B(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2$
$f'(x)=A'(x)B(x)+A(x)B'(x)$ $A'(x)=2$ $B'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}$
$f'(x)=\mathbf {2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}}$
$f(x)=A(x)B(x);\quad A(x)=2x+1;\quad B(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2$
$f'(x)=A'(x)B(x)+A(x)B'(x)$ $A'(x)=2$ $B'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}$
$f'(x)=\mathbf {2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}}$
$f(x)={\frac {2x+1}{\sqrt {2x+2}}}\,$
$f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+1;\quad D(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2$
$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$ $N'(x)=2$ $D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}$
$f'(x)={\frac {{\sqrt {2x+2}}(2)-{\frac {(2x+1)}{\sqrt {2x+2}}}}{2x+2}}=\mathbf {\frac {2x+3}{(2x+2)^{3/2}}}$
$f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+1;\quad D(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2$
$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$ $N'(x)=2$ $D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}$
$f'(x)={\frac {{\sqrt {2x+2}}(2)-{\frac {(2x+1)}{\sqrt {2x+2}}}}{2x+2}}=\mathbf {\frac {2x+3}{(2x+2)^{3/2}}}$
$f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}\,$
$f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x^{2}+1;\quad B(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=3x^{4}+2x$
$f'(x)=A'(x)B(x)+A(x)B'(x)$ $A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x^{2}+1}}}}(4x)={\frac {2x}{\sqrt {2x^{2}+1}}}$ $B'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=2(3x^{4}+2x)(12x^{3}+2)$
$f'(x)={\frac {2x}{\sqrt {2x^{2}+1}}}(3x^{4}+2x)^{2}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)=\mathbf {{\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)}$
$f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x^{2}+1;\quad B(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=3x^{4}+2x$
$f'(x)=A'(x)B(x)+A(x)B'(x)$ $A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x^{2}+1}}}}(4x)={\frac {2x}{\sqrt {2x^{2}+1}}}$ $B'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=2(3x^{4}+2x)(12x^{3}+2)$
$f'(x)={\frac {2x}{\sqrt {2x^{2}+1}}}(3x^{4}+2x)^{2}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)=\mathbf {{\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)}$
$f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}\,$
$f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+3;\quad D(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x^{4}+4x+2$
$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$ $N'(x)=2$ $D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x^{4}+4x+2)(4x^{3}+4)$
$f'(x)={\frac {(x^{4}+4x+2)^{2}(2)-(2x+3)(2)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}=\mathbf {\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}$
$f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+3;\quad D(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x^{4}+4x+2$
$f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}$ $N'(x)=2$ $D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x^{4}+4x+2)(4x^{3}+4)$
$f'(x)={\frac {(x^{4}+4x+2)^{2}(2)-(2x+3)(2)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}=\mathbf {\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}$
$f(x)={\sqrt {x^{3}+1}}(x^{2}-1)\,$
$f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=x^{3}+1;\quad B(x)=x^{2}-1$
$f'(x)=A'(x)B(x)+A(x)B'(x)$ $A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {x^{3}+1}}}}(3x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}$ $B'(x)=2x$
$f'(x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}(x^{2}-1)+{\sqrt {x^{3}+1}}(2x)=\mathbf {{\frac {3x(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}}$
$f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=x^{3}+1;\quad B(x)=x^{2}-1$
$f'(x)=A'(x)B(x)+A(x)B'(x)$ $A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {x^{3}+1}}}}(3x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}$ $B'(x)=2x$
$f'(x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}(x^{2}-1)+{\sqrt {x^{3}+1}}(2x)=\mathbf {{\frac {3x(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}}$
$f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}\,$
$f(x)=g((h(x));\quad g(x)=x^{2};\quad h(x)=A(x)+B(x)+2;\quad A(x)=r(s(x));\quad r(x)=x^{4};\quad s(x)=2x+3;\quad B(x)=4(2x+3)$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(A(x)+B(x)+2)(A'(x)+B'(x))$ $A'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=4(2x+3)^{3}(2)=8(2x+3)^{3}$ $B'(x)=8$
$f'(x)=\mathbf {2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}$
$f(x)=g((h(x));\quad g(x)=x^{2};\quad h(x)=A(x)+B(x)+2;\quad A(x)=r(s(x));\quad r(x)=x^{4};\quad s(x)=2x+3;\quad B(x)=4(2x+3)$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(A(x)+B(x)+2)(A'(x)+B'(x))$ $A'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=4(2x+3)^{3}(2)=8(2x+3)^{3}$ $B'(x)=8$
$f'(x)=\mathbf {2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}$
$f(x)={\sqrt {1+x^{2}}}\,$
$f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}}$
$f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}}$

### Exponentials

$f(x)=(3x^{2}+e)e^{2x}\,$
$f'(x)=f'(x)=(6x)e^{2x}+(3x^{2}+e)2e^{2x}=\mathbf {6xe^{2x}+2e^{2x}(3x^{2}+e)}$
$f'(x)=f'(x)=(6x)e^{2x}+(3x^{2}+e)2e^{2x}=\mathbf {6xe^{2x}+2e^{2x}(3x^{2}+e)}$
$f(x)=e^{2x^{2}+3x}$
$f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}}$
$f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x$
$f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}}$
$f(x)=e^{e^{2x^{2}+1}}$
       $u(x)=e^{x}$
       $v(x)=e^{x}$
       $w(x)=2x^{2}+1$
Then
       $f(x)=u(v(w(x)))$
Using the chain rule, we have
       ${\frac {df}{dx}}={\frac {du}{dv}}{\frac {dv}{dw}}{\frac {dw}{dx}}$
The individual factor are
       ${\frac {du}{dv}}={\frac {d(e^{v})}{dv}}=e^{v}=e^{e^{w}}=e^{e^{2x^{2}+1}}$
       ${\frac {dv}{dw}}={\frac {d(e^{w})}{dw}}=e^{w}=e^{2x^{2}+1}$
       ${\frac {dw}{dx}}={\frac {d(2x^{2}+1)}{dx}}=4x$
So
       ${\frac {df}{dx}}=(e^{e^{2x^{2}+1}})(e^{2x^{2}+1})(4x)=\mathbf {4xe^{2x^{2}+1+e^{2x^{2}+1}}}$
       $u(x)=e^{x}$
       $v(x)=e^{x}$
       $w(x)=2x^{2}+1$
Then
       $f(x)=u(v(w(x)))$
Using the chain rule, we have
       ${\frac {df}{dx}}={\frac {du}{dv}}{\frac {dv}{dw}}{\frac {dw}{dx}}$
The individual factor are
       ${\frac {du}{dv}}={\frac {d(e^{v})}{dv}}=e^{v}=e^{e^{w}}=e^{e^{2x^{2}+1}}$
       ${\frac {dv}{dw}}={\frac {d(e^{w})}{dw}}=e^{w}=e^{2x^{2}+1}$
       ${\frac {dw}{dx}}={\frac {d(2x^{2}+1)}{dx}}=4x$
So
       ${\frac {df}{dx}}=(e^{e^{2x^{2}+1}})(e^{2x^{2}+1})(4x)=\mathbf {4xe^{2x^{2}+1+e^{2x^{2}+1}}}$
$f(x)=4^{x}\,$
$f'(x)=\mathbf {\ln(4)4^{x}}$
$f'(x)=\mathbf {\ln(4)4^{x}}$

### Logarithms

$f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln x\,$
$f'(x)=\ln(2)2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+2^{x-3}{\frac {3}{2{\sqrt {x^{3}-2}}}}3x^{2}+{\frac {1}{x}}=\mathbf {3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-4}}{\sqrt {x^{3}-2}}}+{\frac {1}{x}}}$
$f'(x)=\ln(2)2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+2^{x-3}{\frac {3}{2{\sqrt {x^{3}-2}}}}3x^{2}+{\frac {1}{x}}=\mathbf {3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-4}}{\sqrt {x^{3}-2}}}+{\frac {1}{x}}}$
$f(x)=\ln x-2e^{x}+{\sqrt {x}}\,$
$f'(x)=\mathbf {{\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}}$
$f'(x)=\mathbf {{\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}}$
$f(x)=\ln(\ln(x^{3}(x+1)))\,$
$f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)$
$f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}$
$f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)$
$f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}$
$f(x)=\ln(2x^{2}+3x)\,$
$f'(x)={\frac {1}{2x^{2}+3x}}(4x+3)=\mathbf {\frac {4x+3}{2x^{2}+3x}}$
$f'(x)={\frac {1}{2x^{2}+3x}}(4x+3)=\mathbf {\frac {4x+3}{2x^{2}+3x}}$
$f(x)=\log _{4}x+2\ln x\,$
$f'(x)=\mathbf {{\frac {1}{x\ln 4}}+{\frac {2}{x}}}$
$f'(x)=\mathbf {{\frac {1}{x\ln 4}}+{\frac {2}{x}}}$

### Trigonometric functions

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

## More Differentiation

${\frac {d}{dx}}[(x^{3}+5)^{10}]$
$10(x^{3}+5)^{9}(3x^{2})=\mathbf {30x^{2}(x^{3}+5)^{9}}$
$10(x^{3}+5)^{9}(3x^{2})=\mathbf {30x^{2}(x^{3}+5)^{9}}$
${\frac {d}{dx}}[x^{3}+3x]$
$\mathbf {3x^{2}+3}$
$\mathbf {3x^{2}+3}$
${\frac {d}{dx}}[(x+4)(x+2)(x-3)]$
$f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3$
$f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)$ $A'(x)=B'(x)=C'(x)=1$
$f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}$
$f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3$
$f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)$ $A'(x)=B'(x)=C'(x)=1$
$f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}$
${\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]$
${\frac {3x^{2}-(x+1)(6x)}{(3x^{2})^{2}}}={\frac {3x^{2}-6x^{2}-6x}{9x^{4}}}={\frac {-3x^{2}-6x}{9x^{4}}}=\mathbf {-{\frac {x+2}{3x^{3}}}}$
${\frac {3x^{2}-(x+1)(6x)}{(3x^{2})^{2}}}={\frac {3x^{2}-6x^{2}-6x}{9x^{4}}}={\frac {-3x^{2}-6x}{9x^{4}}}=\mathbf {-{\frac {x+2}{3x^{3}}}}$
${\frac {d}{dx}}[3x^{3}]$
$\mathbf {9x^{2}}$
$\mathbf {9x^{2}}$
${\frac {d}{dx}}[x^{4}\sin x]$
$\mathbf {4x^{3}\sin x+x^{4}\cos x}$
$\mathbf {4x^{3}\sin x+x^{4}\cos x}$
${\frac {d}{dx}}[2^{x}]$
$\mathbf {\ln(2)2^{x}}$
$\mathbf {\ln(2)2^{x}}$
${\frac {d}{dx}}[e^{x^{2}}]$
$\mathbf {2xe^{x^{2}}}$
$\mathbf {2xe^{x^{2}}}$
${\frac {d}{dx}}[e^{2^{x}}]$
$\mathbf {\ln(2)2^{x}e^{2^{x}}}$
$\mathbf {\ln(2)2^{x}e^{2^{x}}}$

## Implicit Differentiation

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

## Logarithmic Differentiation

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

## 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.
$f(x)={\frac {x^{3}}{3}}+x^{2}+5,\;\;\;(3,23)$
$f'(x)=x^{2}+2x$
a) $x^{2}+2x=0\implies \mathbf {x=0,-2}$ b) $m=3^{2}+2(3)=9+6=15$
    $23=15(3)+b\implies b=-22$
    $\mathbf {y=15x-22}$
$f'(x)=x^{2}+2x$
a) $x^{2}+2x=0\implies \mathbf {x=0,-2}$ b) $m=3^{2}+2(3)=9+6=15$
    $23=15(3)+b\implies b=-22$
    $\mathbf {y=15x-22}$
$f(x)=x^{3}-3x+1,\;\;\;(1,-1)$
$f'(x)=3x^{2}-3$
a) $3x^{2}-3=0\implies \mathbf {x=\pm 1}$ b) $m=3(1)^{2}-3=0$
    $-1=b$
    $\mathbf {y=-1}$
$f'(x)=3x^{2}-3$
a) $3x^{2}-3=0\implies \mathbf {x=\pm 1}$ b) $m=3(1)^{2}-3=0$
    $-1=b$
    $\mathbf {y=-1}$
$f(x)={\frac {2}{3}}x^{3}+x^{2}-12x+6,\;\;\;(0,6)$
$f'(x)=2x^{2}+2x-12$
a) $2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf {x=2,-3}$ b) $m=-12$
    $6=b$
    $\mathbf {y=-12x+6}$
$f'(x)=2x^{2}+2x-12$
a) $2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf {x=2,-3}$ b) $m=-12$
    $6=b$
    $\mathbf {y=-12x+6}$
$f(x)=2x+{\frac {1}{\sqrt {x}}},\;\;\;(1,3)$
$f'(x)=2-{\frac {1}{2x^{3/2}}}$
a) $2-{\frac {1}{2x^{3/2}}}=0\implies 2x^{3/2}={\frac {1}{2}}\implies x^{3/2}={\frac {1}{4}}=2^{-2}\implies \mathbf {x=2^{-4/3}}$ b) $m=2-{\frac {1}{2(1)^{3/2}}}=2-{\frac {1}{2}}={\frac {3}{2}}$
    $3={\frac {3}{2}}(1)+b\implies b={\frac {3}{2}}$
    $\mathbf {y={\frac {3}{2}}x+{\frac {3}{2}}}$
$f'(x)=2-{\frac {1}{2x^{3/2}}}$
a) $2-{\frac {1}{2x^{3/2}}}=0\implies 2x^{3/2}={\frac {1}{2}}\implies x^{3/2}={\frac {1}{4}}=2^{-2}\implies \mathbf {x=2^{-4/3}}$ b) $m=2-{\frac {1}{2(1)^{3/2}}}=2-{\frac {1}{2}}={\frac {3}{2}}$
    $3={\frac {3}{2}}(1)+b\implies b={\frac {3}{2}}$
    $\mathbf {y={\frac {3}{2}}x+{\frac {3}{2}}}$
$f(x)=(x^{2}+1)(2-x),\;\;\;(2,0)$
$f'(x)=(2x)(2-x)-(x^{2}+1)=4x-2x^{2}-x^{2}-1=-3x^{2}+4x-1$
a) $-3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf {x=1,{\frac {1}{3}}}$ b) $m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5$
    $0=-5(2)+b\implies b=10$
    $\mathbf {y=-5x+10}$
$f'(x)=(2x)(2-x)-(x^{2}+1)=4x-2x^{2}-x^{2}-1=-3x^{2}+4x-1$
a) $-3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf {x=1,{\frac {1}{3}}}$ b) $m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5$
    $0=-5(2)+b\implies b=10$
    $\mathbf {y=-5x+10}$
$f(x)={\frac {2}{3}}x^{3}+{\frac {5}{2}}x^{2}+2x+1,\;\;\;(3,{\frac {95}{2}})$
$f'(x)=2x^{2}+5x+2$
a) $2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf {x=-{\frac {1}{2}},-2}$ / b) $m=2(3)^{2}+5(3)+2=18+15+2=35$
    ${\frac {95}{2}}=35(3)+b\implies b=-{\frac {115}{2}}$
    $\mathbf {y=35x-{\frac {115}{2}}}$
$f'(x)=2x^{2}+5x+2$
a) $2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf {x=-{\frac {1}{2}},-2}$ / b) $m=2(3)^{2}+5(3)+2=18+15+2=35$
    ${\frac {95}{2}}=35(3)+b\implies b=-{\frac {115}{2}}$
    $\mathbf {y=35x-{\frac {115}{2}}}$
$(x-y-1)^{3}=x\,$
$3(x-y-1)^{2}(1-y')=1$
$1-y'={\frac {1}{3(x-y-1)^{2}}}$ $y'=1-{\frac {1}{3(x-y-1)^{2}}}$ $m=1-{\frac {1}{3(1-(-1)-1)^{2}}}=1-{\frac {1}{3(1)^{2}}}={\frac {2}{3}}$ $-1={\frac {2}{3}}(1)+b\implies b=-{\frac {5}{3}}$
$\mathbf {y={\frac {2}{3}}x-{\frac {5}{3}}}$
$3(x-y-1)^{2}(1-y')=1$
$1-y'={\frac {1}{3(x-y-1)^{2}}}$ $y'=1-{\frac {1}{3(x-y-1)^{2}}}$ $m=1-{\frac {1}{3(1-(-1)-1)^{2}}}=1-{\frac {1}{3(1)^{2}}}={\frac {2}{3}}$ $-1={\frac {2}{3}}(1)+b\implies b=-{\frac {5}{3}}$
$\mathbf {y={\frac {2}{3}}x-{\frac {5}{3}}}$
$e^{xy}+x^{2}=y^{2}\,$
$y'e^{xy}+2x=2yy'$
$y'(e^{xy}-2y)=-2x$ $y'={\frac {2x}{2y-e^{xy}}}$ $m={\frac {2(1)}{2(0)-e^{1(0)}}}={\frac {2}{-1}}=-2$ $0=-2(1)+b\implies b=2$
$\mathbf {y=-2x+2}$
$y'e^{xy}+2x=2yy'$
$y'(e^{xy}-2y)=-2x$ $y'={\frac {2x}{2y-e^{xy}}}$ $m={\frac {2(1)}{2(0)-e^{1(0)}}}={\frac {2}{-1}}=-2$ $0=-2(1)+b\implies b=2$
$\mathbf {y=-2x+2}$

## Higher Order Derivatives

$3x^{4}+3x^{2}+2x$
${\frac {d}{dx}}(3x^{4}+3x^{2}+2x)=12x^{3}+6x+2$
${\frac {d^{2}}{dx^{2}}}(3x^{4}+3x^{2}+2x)={\frac {d}{dx}}(12x^{3}+6x+2)=\mathbf {36x^{2}+6}$
${\frac {d}{dx}}(3x^{4}+3x^{2}+2x)=12x^{3}+6x+2$
${\frac {d^{2}}{dx^{2}}}(3x^{4}+3x^{2}+2x)={\frac {d}{dx}}(12x^{3}+6x+2)=\mathbf {36x^{2}+6}$
$c$
${\frac {dc}{dx}}=0$
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.
${\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0$
$c$
${\frac {dc}{dx}}=0$
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.
${\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0$

## Advanced Understanding of Derivatives

$f^{\prime }(x)$
$f(x)$
$-f(x)$
$-f^{\prime }(x)$
$f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}$
$g(x)=-f(x)$
${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$
$f^{\prime }(x)$
$f(x)$
$-f^{\prime }(x)$
$-f(x)$
$\blacksquare$
$f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}$
$g(x)=-f(x)$
${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$
$f^{\prime }(x)$
$f(x)$
$-f^{\prime }(x)$
$-f(x)$
$\blacksquare$
$f(x)$
$-2\leq x\leq 2$
$f(-2)=f(2)<0$
$f(x)$
    (a) the Intermediate Value Theorem;
    (b) Rolle's Theorem;
    (c) the Extreme Value Theorem.
    (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 $x\in \left[-2,2\right]$.
    (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 $x\in \left[-2,2\right]$.
$g(x)=f^{-1}(x)$
$f^{-1}(x)$
$f(x)$
$f(x)$
$g^{\prime }(x)$
$g^{\prime }(x)$
$g(x)=f^{-1}(x)$
$f\left(g(x)\right)=x$
    ${\begin{aligned}{\frac {d}{dx}}\left[f\left(g(x)\right)\right]&={\frac {d}{dx}}\left(x\right)\\g^{\prime }(x)f^{\prime }\left(g(x)\right)&=1\\g^{\prime }(x)&={\frac {1}{f^{\prime }\left(g(x)\right)}}\\&={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}\\\end{aligned}}$
$g(x)=f^{-1}(x)$
$f\left(g(x)\right)=x$
    ${\begin{aligned}{\frac {d}{dx}}\left[f\left(g(x)\right)\right]&={\frac {d}{dx}}\left(x\right)\\g^{\prime }(x)f^{\prime }\left(g(x)\right)&=1\\g^{\prime }(x)&={\frac {1}{f^{\prime }\left(g(x)\right)}}\\&={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}\\\end{aligned}}$
$f(x)={\begin{cases}2x^{2}+4,&x<-1\\ax^{3}-2x,&x\geq -1\end{cases}}$
$a$
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.
$a=-4$
    $2(-1)^{2}+4=a(-1)^{3}-2(-1)\Rightarrow a=-4$. However, for $f^{\prime }(x)={\begin{cases}4x,&x<-1\\3ax^{2}-2,&x\geq -1\end{cases}}$, we find that $4(-1)\neq -12(-1)^{2}$, so $a=-4$ makes the function continuous but non-differentiable.
(b) There is no $a\in \mathbb {R}$ that allows the function to be differentiable and continuous.
    A proof of this is simple.
       ${\begin{aligned}\lim _{h\to 0^{+}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{+}}{\frac {2(-1+h)^{2}+4-4-2}{h}}\\&=\lim _{h\to 0^{+}}{\frac {-2-4h+2h^{2}-2}{h}}\\&=-4\\\end{aligned}}$
    However,
       ${\begin{aligned}\lim _{h\to 0^{-}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{-}}{\frac {a(-1+h)^{3}+-2(-1+h)-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a\left(1+3h-3h^{2}+h^{3}\right)+2-2h-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a+3ah-3ah^{2}+ah^{3}-2h-4}{h}}\\&=\lim _{h\to 0^{-}}{\frac {3ah-3ah^{2}+ah^{3}-2h+(a-4)}{h}}\end{aligned}}$
    To allow the best possible chance, we will let $a=4$:
       ${\begin{aligned}\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h+(4-4)}{h}}&=\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h}{h}}\\&=\lim _{h\to 0^{-}}12-12h+4h^{2}-2\\&=10\end{aligned}}$
    For any other $a\in \mathbb {R}$, one will have an infinity on the left-hand sided limit. Therefore, there is no possible $a\in \mathbb {R}$ that allows the function to be differentiable and continuous.
$a=-4$
    $2(-1)^{2}+4=a(-1)^{3}-2(-1)\Rightarrow a=-4$. However, for $f^{\prime }(x)={\begin{cases}4x,&x<-1\\3ax^{2}-2,&x\geq -1\end{cases}}$, we find that $4(-1)\neq -12(-1)^{2}$, so $a=-4$ makes the function continuous but non-differentiable.
(b) There is no $a\in \mathbb {R}$ that allows the function to be differentiable and continuous.
    A proof of this is simple.
       ${\begin{aligned}\lim _{h\to 0^{+}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{+}}{\frac {2(-1+h)^{2}+4-4-2}{h}}\\&=\lim _{h\to 0^{+}}{\frac {-2-4h+2h^{2}-2}{h}}\\&=-4\\\end{aligned}}$
    However,
       ${\begin{aligned}\lim _{h\to 0^{-}}{\frac {f(-1+h)-f(-1)}{h}}&=\lim _{h\to 0^{-}}{\frac {a(-1+h)^{3}+-2(-1+h)-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a\left(1+3h-3h^{2}+h^{3}\right)+2-2h-6}{h}}\\&=\lim _{h\to 0^{-}}{\frac {a+3ah-3ah^{2}+ah^{3}-2h-4}{h}}\\&=\lim _{h\to 0^{-}}{\frac {3ah-3ah^{2}+ah^{3}-2h+(a-4)}{h}}\end{aligned}}$
    To allow the best possible chance, we will let $a=4$:
       ${\begin{aligned}\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h+(4-4)}{h}}&=\lim _{h\to 0^{-}}{\frac {3(4)h-3(4)h^{2}+(4)h^{3}-2h}{h}}\\&=\lim _{h\to 0^{-}}12-12h+4h^{2}-2\\&=10\end{aligned}}$
    For any other $a\in \mathbb {R}$, one will have an infinity on the left-hand sided limit. Therefore, there is no possible $a\in \mathbb {R}$ that allows the function to be differentiable and continuous.
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