# Calculus/Differentiation/Basics of Differentiation/Solutions

## Find the Derivative by Definition

1. ${\displaystyle f(x)=x^{2}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
2. ${\displaystyle f(x)=2x+2\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
3. ${\displaystyle f(x)={\frac {1}{2}}x^{2}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
4. ${\displaystyle f(x)=2x^{2}+4x+4\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
5. ${\displaystyle f(x)={\sqrt {x+2}}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
6. ${\displaystyle f(x)={\frac {1}{x}}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
7. ${\displaystyle f(x)={\frac {3}{x+1}}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
8. ${\displaystyle f(x)={\frac {1}{\sqrt {x+1}}}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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}}}
9. ${\displaystyle f(x)={\frac {x}{x+2}}\,}$
{\displaystyle {\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}}}
{\displaystyle {\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

10. Use the definition of the derivative to prove that for any fixed real number ${\displaystyle c}$, ${\displaystyle {\frac {d}{dx}}\left[cf(x)\right]=c{\frac {d}{dx}}\left[f(x)\right]}$
{\displaystyle {\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}}}
{\displaystyle {\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

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

### Product Rule

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

### Quotient Rule

29. ${\displaystyle f(x)={\frac {2x+1}{x+5}}\,}$
${\displaystyle f'(x)={\frac {(x+5)(2)-(2x+1)}{(x+5)^{2}}}=\mathbf {\frac {9}{(x+5)^{2}}} }$
${\displaystyle f'(x)={\frac {(x+5)(2)-(2x+1)}{(x+5)^{2}}}=\mathbf {\frac {9}{(x+5)^{2}}} }$
30. ${\displaystyle f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}\,}$
${\displaystyle f'(x)={\frac {(3x^{2}+1)(12x^{3}+2)-(3x^{4}+2x+2)(6x)}{(3x^{2}+1)^{2}}}=\mathbf {\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}} }$
${\displaystyle f'(x)={\frac {(3x^{2}+1)(12x^{3}+2)-(3x^{4}+2x+2)(6x)}{(3x^{2}+1)^{2}}}=\mathbf {\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}} }$
31. ${\displaystyle f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}\,}$
${\displaystyle f'(x)={\frac {(x+2)({\frac {3}{2}}{\sqrt {x}})-(x^{\frac {3}{2}}+1)}{(x+2)^{2}}}=\mathbf {\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}} }$
${\displaystyle f'(x)={\frac {(x+2)({\frac {3}{2}}{\sqrt {x}})-(x^{\frac {3}{2}}+1)}{(x+2)^{2}}}=\mathbf {\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}} }$
32. ${\displaystyle d(u)={\frac {u^{3}+2}{u^{3}}}\,}$
${\displaystyle d'(u)={\frac {u^{3}(3u^{2})-(u^{3}+2)(3u^{2})}{u^{6}}}=\mathbf {-{\frac {6}{u^{4}}}} }$
${\displaystyle d'(u)={\frac {u^{3}(3u^{2})-(u^{3}+2)(3u^{2})}{u^{6}}}=\mathbf {-{\frac {6}{u^{4}}}} }$
33. ${\displaystyle f(x)={\frac {x^{2}+x}{2x-1}}\,}$
${\displaystyle f'(x)={\frac {(2x-1)(2x+1)-(x^{2}+x)(2)}{(2x-1)^{2}}}=\mathbf {\frac {2x^{2}-2x-1}{(2x-1)^{2}}} }$
${\displaystyle f'(x)={\frac {(2x-1)(2x+1)-(x^{2}+x)(2)}{(2x-1)^{2}}}=\mathbf {\frac {2x^{2}-2x-1}{(2x-1)^{2}}} }$
34. ${\displaystyle f(x)={\frac {x+1}{2x^{2}+2x+3}}\,}$
${\displaystyle f'(x)={\frac {(2x^{2}+2x+3)-(x+1)(4x+2)}{(2x^{2}+2x+3)^{2}}}=\mathbf {\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}} }$
${\displaystyle f'(x)={\frac {(2x^{2}+2x+3)-(x+1)(4x+2)}{(2x^{2}+2x+3)^{2}}}=\mathbf {\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}} }$
35. ${\displaystyle f(x)={\frac {16x^{4}+2x^{2}}{x}}\,}$
${\displaystyle f'(x)={\frac {x(64x^{3}+4x)-(16x^{4}+2x^{2})}{x^{2}}}=\mathbf {48x^{2}+2} }$
${\displaystyle f'(x)={\frac {x(64x^{3}+4x)-(16x^{4}+2x^{2})}{x^{2}}}=\mathbf {48x^{2}+2} }$

### Chain Rule

43. ${\displaystyle f(x)=(x+5)^{2}\,}$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)} }$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x+5}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x+5)(1)=\mathbf {2(x+5)} }$
44. ${\displaystyle g(x)=(x^{3}-2x+5)^{2}\,}$
Let ${\displaystyle g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5}$. Then
${\displaystyle g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)} }$
Let ${\displaystyle g(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=x^{3}-2x+5}$. Then
${\displaystyle g'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=\mathbf {2(x^{3}-2x+5)(3x^{2}-2)} }$
45. ${\displaystyle f(x)={\sqrt {1-x^{2}}}\,}$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}} }$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1-x^{2}}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1-x^{2}}}}}(-2x)=\mathbf {-{\frac {x}{\sqrt {1-x^{2}}}}} }$
46. ${\displaystyle f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}\,}$
Let ${\displaystyle f(x)={\frac {N(x)}{D(x)}};\quad N(x)=g(h(x));\quad g(x)=x^{3};\quad h(x)=2x+4;\quad D(x)=4x^{3}+1}$. Then

${\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}}$
${\displaystyle N'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=3(2x+4)^{2}(2)=6(2x+4)^{2}}$
${\displaystyle D'(x)=12x^{2}}$

${\displaystyle f'(x)={\frac {(4x^{3}+1)6(2x+4)^{2}-(2x+4)^{3}12x^{2}}{(4x^{3}+1)^{2}}}=\mathbf {\frac {6(4x^{3}+1)(2x+4)^{2}-12x^{2}(2x+4)^{3}}{(4x^{3}+1)^{2}}} }$
Let ${\displaystyle f(x)={\frac {N(x)}{D(x)}};\quad N(x)=g(h(x));\quad g(x)=x^{3};\quad h(x)=2x+4;\quad D(x)=4x^{3}+1}$. Then

${\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}}$
${\displaystyle N'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=3(2x+4)^{2}(2)=6(2x+4)^{2}}$
${\displaystyle D'(x)=12x^{2}}$

${\displaystyle f'(x)={\frac {(4x^{3}+1)6(2x+4)^{2}-(2x+4)^{3}12x^{2}}{(4x^{3}+1)^{2}}}=\mathbf {\frac {6(4x^{3}+1)(2x+4)^{2}-12x^{2}(2x+4)^{3}}{(4x^{3}+1)^{2}}} }$
47. ${\displaystyle f(x)=(2x+1){\sqrt {2x+2}}\,}$
Let ${\displaystyle f(x)=A(x)B(x);\quad A(x)=2x+1;\quad B(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2}$. Then

${\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)}$
${\displaystyle A'(x)=2}$
${\displaystyle B'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}}$

${\displaystyle f'(x)=\mathbf {2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}} }$
Let ${\displaystyle f(x)=A(x)B(x);\quad A(x)=2x+1;\quad B(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2}$. Then

${\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)}$
${\displaystyle A'(x)=2}$
${\displaystyle B'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}}$

${\displaystyle f'(x)=\mathbf {2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}} }$
48. ${\displaystyle f(x)={\frac {2x+1}{\sqrt {2x+2}}}\,}$
Let ${\displaystyle f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+1;\quad D(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2}$. Then

${\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}}$
${\displaystyle N'(x)=2}$
${\displaystyle D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}}$

${\displaystyle f'(x)={\frac {{\sqrt {2x+2}}(2)-{\frac {(2x+1)}{\sqrt {2x+2}}}}{2x+2}}=\mathbf {\frac {2x+3}{(2x+2)^{3/2}}} }$
Let ${\displaystyle f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+1;\quad D(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x+2}$. Then

${\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}}$
${\displaystyle N'(x)=2}$
${\displaystyle D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x+2}}}}(2)={\frac {1}{\sqrt {2x+2}}}}$

${\displaystyle f'(x)={\frac {{\sqrt {2x+2}}(2)-{\frac {(2x+1)}{\sqrt {2x+2}}}}{2x+2}}=\mathbf {\frac {2x+3}{(2x+2)^{3/2}}} }$
49. ${\displaystyle f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}\,}$
Let ${\displaystyle f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x^{2}+1;\quad B(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=3x^{4}+2x}$. Then

${\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)}$
${\displaystyle A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x^{2}+1}}}}(4x)={\frac {2x}{\sqrt {2x^{2}+1}}}}$
${\displaystyle B'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=2(3x^{4}+2x)(12x^{3}+2)}$

${\displaystyle f'(x)={\frac {2x}{\sqrt {2x^{2}+1}}}(3x^{4}+2x)^{2}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)=\mathbf {{\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)} }$
Let ${\displaystyle f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=2x^{2}+1;\quad B(x)=r(s(x));\quad r(x)=x^{2};\quad s(x)=3x^{4}+2x}$. Then

${\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)}$
${\displaystyle A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {2x^{2}+1}}}}(4x)={\frac {2x}{\sqrt {2x^{2}+1}}}}$
${\displaystyle B'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=2(3x^{4}+2x)(12x^{3}+2)}$

${\displaystyle f'(x)={\frac {2x}{\sqrt {2x^{2}+1}}}(3x^{4}+2x)^{2}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)=\mathbf {{\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)} }$
50. ${\displaystyle f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}\,}$
Let ${\displaystyle f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+3;\quad D(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x^{4}+4x+2}$. Then

${\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}}$
${\displaystyle N'(x)=2}$
${\displaystyle D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x^{4}+4x+2)(4x^{3}+4)}$

${\displaystyle f'(x)={\frac {(x^{4}+4x+2)^{2}(2)-(2x+3)(2)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}=\mathbf {\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}} }$
Let ${\displaystyle f(x)={\frac {N(x)}{D(x)}};\quad N(x)=2x+3;\quad D(x)=g(h(x));\quad g(x)=x^{2};\quad h(x)=x^{4}+4x+2}$. Then

${\displaystyle f'(x)={\frac {D(x)N'(x)-N(x)D'(x)}{D^{2}(x)}}}$
${\displaystyle N'(x)=2}$
${\displaystyle D'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(x^{4}+4x+2)(4x^{3}+4)}$

${\displaystyle f'(x)={\frac {(x^{4}+4x+2)^{2}(2)-(2x+3)(2)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}=\mathbf {\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}} }$
51. ${\displaystyle f(x)={\sqrt {x^{3}+1}}(x^{2}-1)\,}$
Let ${\displaystyle f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=x^{3}+1;\quad B(x)=x^{2}-1}$. Then

${\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)}$
${\displaystyle A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {x^{3}+1}}}}(3x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}}$
${\displaystyle B'(x)=2x}$

${\displaystyle f'(x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}(x^{2}-1)+{\sqrt {x^{3}+1}}(2x)=\mathbf {{\frac {3x(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}} }$
Let ${\displaystyle f(x)=A(x)B(x);\quad A(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=x^{3}+1;\quad B(x)=x^{2}-1}$. Then

${\displaystyle f'(x)=A'(x)B(x)+A(x)B'(x)}$
${\displaystyle A'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {x^{3}+1}}}}(3x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}}$
${\displaystyle B'(x)=2x}$

${\displaystyle f'(x)={\frac {3x}{2{\sqrt {x^{3}+1}}}}(x^{2}-1)+{\sqrt {x^{3}+1}}(2x)=\mathbf {{\frac {3x(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}} }$
52. ${\displaystyle f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}\,}$
Let ${\displaystyle f(x)=g((h(x));\quad g(x)=x^{2};\quad h(x)=A(x)+B(x)+2;\quad A(x)=r(s(x));\quad r(x)=x^{4};\quad s(x)=2x+3;\quad B(x)=4(2x+3)}$. Then

${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(A(x)+B(x)+2)(A'(x)+B'(x))}$
${\displaystyle A'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=4(2x+3)^{3}(2)=8(2x+3)^{3}}$
${\displaystyle B'(x)=8}$

${\displaystyle f'(x)=\mathbf {2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)} }$
Let ${\displaystyle f(x)=g((h(x));\quad g(x)=x^{2};\quad h(x)=A(x)+B(x)+2;\quad A(x)=r(s(x));\quad r(x)=x^{4};\quad s(x)=2x+3;\quad B(x)=4(2x+3)}$. Then

${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=2(A(x)+B(x)+2)(A'(x)+B'(x))}$
${\displaystyle A'(x)={\frac {dr}{ds}}{\frac {ds}{dx}}=4(2x+3)^{3}(2)=8(2x+3)^{3}}$
${\displaystyle B'(x)=8}$

${\displaystyle f'(x)=\mathbf {2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)} }$
53. ${\displaystyle f(x)={\sqrt {1+x^{2}}}\,}$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)={\sqrt {x}};\quad h(x)=1+x^{2}}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}={\frac {1}{2{\sqrt {1+x^{2}}}}}(2x)=\mathbf {\frac {x}{\sqrt {1+x^{2}}}} }$

### Exponentials

54. ${\displaystyle f(x)=(3x^{2}+e)e^{2x}\,}$
${\displaystyle f'(x)=f'(x)=(6x)e^{2x}+(3x^{2}+e)2e^{2x}=\mathbf {6xe^{2x}+2e^{2x}(3x^{2}+e)} }$
${\displaystyle f'(x)=f'(x)=(6x)e^{2x}+(3x^{2}+e)2e^{2x}=\mathbf {6xe^{2x}+2e^{2x}(3x^{2}+e)} }$
55. ${\displaystyle f(x)=e^{2x^{2}+3x}}$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}} }$
Let ${\displaystyle f(x)=g(h(x));\quad g(x)=e^{x};\quad h(x)=2x^{2}+3x}$. Then
${\displaystyle f'(x)={\frac {dg}{dh}}{\frac {dh}{dx}}=e^{2x^{2}+3x}(4x+3)=\mathbf {(4x+3)e^{2x^{2}+3x}} }$
56. ${\displaystyle f(x)=e^{e^{2x^{2}+1}}}$
Let
${\displaystyle u(x)=e^{x}}$
${\displaystyle v(x)=e^{x}}$
${\displaystyle w(x)=2x^{2}+1}$

Then

${\displaystyle f(x)=u(v(w(x)))}$

Using the chain rule, we have

${\displaystyle {\frac {df}{dx}}={\frac {du}{dv}}{\frac {dv}{dw}}{\frac {dw}{dx}}}$

The individual factor are

${\displaystyle {\frac {du}{dv}}={\frac {d(e^{v})}{dv}}=e^{v}=e^{e^{w}}=e^{e^{2x^{2}+1}}}$
${\displaystyle {\frac {dv}{dw}}={\frac {d(e^{w})}{dw}}=e^{w}=e^{2x^{2}+1}}$
${\displaystyle {\frac {dw}{dx}}={\frac {d(2x^{2}+1)}{dx}}=4x}$

So

${\displaystyle {\frac {df}{dx}}=(e^{e^{2x^{2}+1}})(e^{2x^{2}+1})(4x)=\mathbf {4xe^{2x^{2}+1+e^{2x^{2}+1}}} }$
Let
${\displaystyle u(x)=e^{x}}$
${\displaystyle v(x)=e^{x}}$
${\displaystyle w(x)=2x^{2}+1}$

Then

${\displaystyle f(x)=u(v(w(x)))}$

Using the chain rule, we have

${\displaystyle {\frac {df}{dx}}={\frac {du}{dv}}{\frac {dv}{dw}}{\frac {dw}{dx}}}$

The individual factor are

${\displaystyle {\frac {du}{dv}}={\frac {d(e^{v})}{dv}}=e^{v}=e^{e^{w}}=e^{e^{2x^{2}+1}}}$
${\displaystyle {\frac {dv}{dw}}={\frac {d(e^{w})}{dw}}=e^{w}=e^{2x^{2}+1}}$
${\displaystyle {\frac {dw}{dx}}={\frac {d(2x^{2}+1)}{dx}}=4x}$

So

${\displaystyle {\frac {df}{dx}}=(e^{e^{2x^{2}+1}})(e^{2x^{2}+1})(4x)=\mathbf {4xe^{2x^{2}+1+e^{2x^{2}+1}}} }$
57. ${\displaystyle f(x)=4^{x}\,}$
${\displaystyle f'(x)=\mathbf {\ln(4)4^{x}} }$
${\displaystyle f'(x)=\mathbf {\ln(4)4^{x}} }$

### Logarithms

58. ${\displaystyle f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln x\,}$
${\displaystyle f'(x)=\ln(2)2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+2^{x-3}{\frac {3}{2{\sqrt {x^{3}-2}}}}3x^{2}+{\frac {1}{x}}=\mathbf {3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-4}}{\sqrt {x^{3}-2}}}+{\frac {1}{x}}} }$
${\displaystyle f'(x)=\ln(2)2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+2^{x-3}{\frac {3}{2{\sqrt {x^{3}-2}}}}3x^{2}+{\frac {1}{x}}=\mathbf {3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-4}}{\sqrt {x^{3}-2}}}+{\frac {1}{x}}} }$
47. ${\displaystyle f(x)=\ln x-2e^{x}+{\sqrt {x}}\,}$
${\displaystyle f'(x)=\mathbf {{\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}} }$
${\displaystyle f'(x)=\mathbf {{\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}} }$
59. ${\displaystyle f(x)=\ln(\ln(x^{3}(x+1)))\,}$
Let ${\displaystyle f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)}$. Then
${\displaystyle f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}} }$
Let ${\displaystyle f(x)=g(g(h(x)));\quad g(x)=\ln(x);\quad h(x)=x^{3}(x+1)}$. Then
${\displaystyle f'(x)={\frac {dg(g(h(x)))}{dg(h(x))}}{\frac {dg(h(x))}{dh(x)}}{\frac {dh(x)}{dx}}={\frac {1}{\ln(x^{3}(x+1))}}{\frac {1}{x^{3}(x+1)}}(4x^{3}+3x^{2})=\mathbf {\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}} }$
60. ${\displaystyle f(x)=\ln(2x^{2}+3x)\,}$
${\displaystyle f'(x)={\frac {1}{2x^{2}+3x}}(4x+3)=\mathbf {\frac {4x+3}{2x^{2}+3x}} }$
${\displaystyle f'(x)={\frac {1}{2x^{2}+3x}}(4x+3)=\mathbf {\frac {4x+3}{2x^{2}+3x}} }$
61. ${\displaystyle f(x)=\log _{4}x+2\ln x\,}$
${\displaystyle f'(x)=\mathbf {{\frac {1}{x\ln 4}}+{\frac {2}{x}}} }$
${\displaystyle f'(x)=\mathbf {{\frac {1}{x\ln 4}}+{\frac {2}{x}}} }$

### Trigonometric functions

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

## More Differentiation

65. ${\displaystyle {\frac {d}{dx}}[(x^{3}+5)^{10}]}$
${\displaystyle 10(x^{3}+5)^{9}(3x^{2})=\mathbf {30x^{2}(x^{3}+5)^{9}} }$
${\displaystyle 10(x^{3}+5)^{9}(3x^{2})=\mathbf {30x^{2}(x^{3}+5)^{9}} }$
66. ${\displaystyle {\frac {d}{dx}}[x^{3}+3x]}$
${\displaystyle \mathbf {3x^{2}+3} }$
${\displaystyle \mathbf {3x^{2}+3} }$
67. ${\displaystyle {\frac {d}{dx}}[(x+4)(x+2)(x-3)]}$
Let ${\displaystyle f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3}$. Then

${\displaystyle f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)}$
${\displaystyle A'(x)=B'(x)=C'(x)=1}$

${\displaystyle f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)} }$
Let ${\displaystyle f(x)=A(x)B(x)C(x);\quad A(x)=x+4;\quad B(x)=x+2;\quad C(x)=x-3}$. Then

${\displaystyle f'(x)=A'(x)B(x)C(x)+A(x)B'(x)C(x)+A(x)B(x)C'(x)}$
${\displaystyle A'(x)=B'(x)=C'(x)=1}$

${\displaystyle f'(x)=\mathbf {(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)} }$
68. ${\displaystyle {\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]}$
${\displaystyle {\frac {3x^{2}-(x+1)(6x)}{(3x^{2})^{2}}}={\frac {3x^{2}-6x^{2}-6x}{9x^{4}}}={\frac {-3x^{2}-6x}{9x^{4}}}=\mathbf {-{\frac {x+2}{3x^{3}}}} }$
${\displaystyle {\frac {3x^{2}-(x+1)(6x)}{(3x^{2})^{2}}}={\frac {3x^{2}-6x^{2}-6x}{9x^{4}}}={\frac {-3x^{2}-6x}{9x^{4}}}=\mathbf {-{\frac {x+2}{3x^{3}}}} }$
69. ${\displaystyle {\frac {d}{dx}}[3x^{3}]}$
${\displaystyle \mathbf {9x^{2}} }$
${\displaystyle \mathbf {9x^{2}} }$
70. ${\displaystyle {\frac {d}{dx}}[x^{4}\sin x]}$
${\displaystyle \mathbf {4x^{3}\sin x+x^{4}\cos x} }$
${\displaystyle \mathbf {4x^{3}\sin x+x^{4}\cos x} }$
71. ${\displaystyle 2^{x}}$
${\displaystyle \mathbf {\ln(2)2^{x}} }$
${\displaystyle \mathbf {\ln(2)2^{x}} }$
72. ${\displaystyle {\frac {d}{dx}}[e^{x^{2}}]}$
${\displaystyle \mathbf {2xe^{x^{2}}} }$
${\displaystyle \mathbf {2xe^{x^{2}}} }$
73. ${\displaystyle {\frac {d}{dx}}[e^{2^{x}}]}$
${\displaystyle \mathbf {\ln(2)2^{x}e^{2^{x}}} }$
${\displaystyle \mathbf {\ln(2)2^{x}e^{2^{x}}} }$

## Implicit Differentiation

Use implicit differentiation to find y'

74. ${\displaystyle x^{3}+y^{3}=xy\,}$
${\displaystyle 3x^{2}+3y^{2}y'=y+xy'}$

${\displaystyle 3y^{2}y'-xy'=y-3x^{2}}$

${\displaystyle \mathbf {y'={\frac {y-3x^{2}}{3y^{2}-x}}} }$
${\displaystyle 3x^{2}+3y^{2}y'=y+xy'}$

${\displaystyle 3y^{2}y'-xy'=y-3x^{2}}$

${\displaystyle \mathbf {y'={\frac {y-3x^{2}}{3y^{2}-x}}} }$
75. ${\displaystyle (2x+y)^{4}+3x^{2}+3y^{2}={\frac {x}{y}}+1\,}$
${\displaystyle 4(2x+y)^{3}(2+y')+6x+6yy'={\frac {y-xy'}{y^{2}}}}$

${\displaystyle 8(2x+y)^{3}+4y'(2x+y)^{3}+6x+6yy'={\frac {1}{y}}-{\frac {xy'}{y^{2}}}}$
${\displaystyle y'(4(2x+y)^{3}+6y+{\frac {x}{y^{2}}})={\frac {1}{y}}-8(2x+y)^{3}-6x}$
${\displaystyle y'({\frac {4y^{2}(2x+y)^{3}+6y^{3}+x}{y^{2}}})={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{y^{2}}}}$

${\displaystyle \mathbf {y'={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}} }$
${\displaystyle 4(2x+y)^{3}(2+y')+6x+6yy'={\frac {y-xy'}{y^{2}}}}$

${\displaystyle 8(2x+y)^{3}+4y'(2x+y)^{3}+6x+6yy'={\frac {1}{y}}-{\frac {xy'}{y^{2}}}}$
${\displaystyle y'(4(2x+y)^{3}+6y+{\frac {x}{y^{2}}})={\frac {1}{y}}-8(2x+y)^{3}-6x}$
${\displaystyle y'({\frac {4y^{2}(2x+y)^{3}+6y^{3}+x}{y^{2}}})={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{y^{2}}}}$

${\displaystyle \mathbf {y'={\frac {y-8y^{2}(2x+y)^{3}-6xy^{2}}{4y^{2}(2x+y)^{3}+6y^{3}+x}}} }$

## Logarithmic Differentiation

Use logarithmic differentiation to find ${\displaystyle {\frac {dy}{dx}}}$:

76. ${\displaystyle y=x({\sqrt[{4}]{1-x^{3}}}\,)}$
${\displaystyle \ln y=\ln(x)+\ln({\sqrt[{4}]{1-x^{3}}})=\ln(x)+{\frac {\ln(1-x^{3})}{4}}}$

${\displaystyle {\frac {y'}{y}}={\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}}}$

${\displaystyle y'=x({\sqrt[{4}]{1-x^{3}}}\,)({\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}})=\mathbf {{\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{3/4}}}} }$
${\displaystyle \ln y=\ln(x)+\ln({\sqrt[{4}]{1-x^{3}}})=\ln(x)+{\frac {\ln(1-x^{3})}{4}}}$

${\displaystyle {\frac {y'}{y}}={\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}}}$

${\displaystyle y'=x({\sqrt[{4}]{1-x^{3}}}\,)({\frac {1}{x}}-{\frac {3x^{2}}{4(1-x^{3})}})=\mathbf {{\sqrt[{4}]{1-x^{3}}}-{\frac {3x^{3}}{4(1-x^{3})^{3/4}}}} }$
77. ${\displaystyle y={\sqrt {x+1 \over 1-x}}\,}$
${\displaystyle \ln y={\frac {1}{2}}(\ln(x+1)-\ln(1-x))}$

${\displaystyle {\frac {y'}{y}}={\frac {1}{2}}({\frac {1}{x+1}}+{\frac {1}{1-x}})}$

${\displaystyle \mathbf {y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}({\frac {1}{x+1}}+{\frac {1}{1-x}})} }$
${\displaystyle \ln y={\frac {1}{2}}(\ln(x+1)-\ln(1-x))}$

${\displaystyle {\frac {y'}{y}}={\frac {1}{2}}({\frac {1}{x+1}}+{\frac {1}{1-x}})}$

${\displaystyle \mathbf {y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}({\frac {1}{x+1}}+{\frac {1}{1-x}})} }$
78. ${\displaystyle y=(2x)^{2x}\,}$
${\displaystyle \ln y=2x\ln(2x)}$

${\displaystyle {\frac {y'}{y}}=2\ln(2x)+2x{\frac {2}{2x}}=2\ln(2x)+2}$

${\displaystyle \mathbf {y'=(2x)^{2x}(2\ln(2x)+2)} }$
${\displaystyle \ln y=2x\ln(2x)}$

${\displaystyle {\frac {y'}{y}}=2\ln(2x)+2x{\frac {2}{2x}}=2\ln(2x)+2}$

${\displaystyle \mathbf {y'=(2x)^{2x}(2\ln(2x)+2)} }$
79. ${\displaystyle y=(x^{3}+4x)^{3x+1}\,}$
${\displaystyle \ln y=(3x+1)\ln(x^{3}+4x)}$

${\displaystyle {\frac {y'}{y}}=3\ln(x^{3}+4x)+(3x+1){\frac {3x^{2}+4}{x^{3}+4x}}}$

${\displaystyle \mathbf {y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})} }$
${\displaystyle \ln y=(3x+1)\ln(x^{3}+4x)}$

${\displaystyle {\frac {y'}{y}}=3\ln(x^{3}+4x)+(3x+1){\frac {3x^{2}+4}{x^{3}+4x}}}$

${\displaystyle \mathbf {y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})} }$
80. ${\displaystyle y=(6x)^{\cos(x)+1}\,}$
${\displaystyle \ln y=(\cos(x)+1)\ln(6x)}$

${\displaystyle {\frac {y'}{y}}=-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}}}$

${\displaystyle \mathbf {y'=(6x)^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})} }$
${\displaystyle \ln y=(\cos(x)+1)\ln(6x)}$

${\displaystyle {\frac {y'}{y}}=-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}}}$

${\displaystyle \mathbf {y'=(6x)^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})} }$

## Equation of Tangent Line

For each function, ${\displaystyle f}$, (a) determine for what values of ${\displaystyle x}$ the tangent line to ${\displaystyle f}$ is horizontal and (b) find an equation of the tangent line to ${\displaystyle f}$ at the given point.

81. ${\displaystyle f(x)={\frac {x^{3}}{3}}+x^{2}+5,\;\;\;(3,23)}$
:${\displaystyle f'(x)=x^{2}+2x}$

a) ${\displaystyle x^{2}+2x=0\implies \mathbf {x=0,-2} }$
b) ${\displaystyle m=3^{2}+2(3)=9+6=15}$

${\displaystyle 23=15(3)+b\implies b=-22}$
${\displaystyle \mathbf {y=15x-22} }$
:${\displaystyle f'(x)=x^{2}+2x}$

a) ${\displaystyle x^{2}+2x=0\implies \mathbf {x=0,-2} }$
b) ${\displaystyle m=3^{2}+2(3)=9+6=15}$

${\displaystyle 23=15(3)+b\implies b=-22}$
${\displaystyle \mathbf {y=15x-22} }$
82. ${\displaystyle f(x)=x^{3}-3x+1,\;\;\;(1,-1)}$
:${\displaystyle f'(x)=3x^{2}-3}$

a) ${\displaystyle 3x^{2}-3=0\implies \mathbf {x=\pm 1} }$
b) ${\displaystyle m=3(1)^{2}-3=0}$

${\displaystyle -1=b}$
${\displaystyle \mathbf {y=-1} }$
:${\displaystyle f'(x)=3x^{2}-3}$

a) ${\displaystyle 3x^{2}-3=0\implies \mathbf {x=\pm 1} }$
b) ${\displaystyle m=3(1)^{2}-3=0}$

${\displaystyle -1=b}$
${\displaystyle \mathbf {y=-1} }$
83. ${\displaystyle f(x)={\frac {2}{3}}x^{3}+x^{2}-12x+6,\;\;\;(0,6)}$
:${\displaystyle f'(x)=2x^{2}+2x-12}$

a) ${\displaystyle 2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf {x=2,-3} }$
b) ${\displaystyle m=-12}$

${\displaystyle 6=b}$
${\displaystyle \mathbf {y=-12x+6} }$
:${\displaystyle f'(x)=2x^{2}+2x-12}$

a) ${\displaystyle 2x^{2}+2x-12=0=x^{2}+x-6=(x+3)(x-2)\implies \mathbf {x=2,-3} }$
b) ${\displaystyle m=-12}$

${\displaystyle 6=b}$
${\displaystyle \mathbf {y=-12x+6} }$
84. ${\displaystyle f(x)=2x+{\frac {1}{\sqrt {x}}},\;\;\;(1,3)}$
:${\displaystyle f'(x)=2-{\frac {1}{2x^{3/2}}}}$

a) ${\displaystyle 2-{\frac {1}{2x^{3/2}}}=0\implies 2x^{3/2}={\frac {1}{2}}\implies x^{3/2}={\frac {1}{4}}=2^{-2}\implies \mathbf {x=2^{-4/3}} }$
b) ${\displaystyle m=2-{\frac {1}{2(1)^{3/2}}}=2-{\frac {1}{2}}={\frac {3}{2}}}$

${\displaystyle 3={\frac {3}{2}}(1)+b\implies b={\frac {3}{2}}}$
${\displaystyle \mathbf {y={\frac {3}{2}}x+{\frac {3}{2}}} }$
:${\displaystyle f'(x)=2-{\frac {1}{2x^{3/2}}}}$

a) ${\displaystyle 2-{\frac {1}{2x^{3/2}}}=0\implies 2x^{3/2}={\frac {1}{2}}\implies x^{3/2}={\frac {1}{4}}=2^{-2}\implies \mathbf {x=2^{-4/3}} }$
b) ${\displaystyle m=2-{\frac {1}{2(1)^{3/2}}}=2-{\frac {1}{2}}={\frac {3}{2}}}$

${\displaystyle 3={\frac {3}{2}}(1)+b\implies b={\frac {3}{2}}}$
${\displaystyle \mathbf {y={\frac {3}{2}}x+{\frac {3}{2}}} }$
85. ${\displaystyle f(x)=(x^{2}+1)(2-x),\;\;\;(2,0)}$
:${\displaystyle f'(x)=(2x)(2-x)-(x^{2}+1)=4x-2x^{2}-x^{2}-1=-3x^{2}+4x-1}$

a) ${\displaystyle -3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf {x=1,{\frac {1}{3}}} }$
b) ${\displaystyle m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5}$

${\displaystyle 0=-5(2)+b\implies b=10}$
${\displaystyle \mathbf {y=-5x+10} }$
:${\displaystyle f'(x)=(2x)(2-x)-(x^{2}+1)=4x-2x^{2}-x^{2}-1=-3x^{2}+4x-1}$

a) ${\displaystyle -3x^{2}+4x-1=0=(3x-1)(-x+1)\implies \mathbf {x=1,{\frac {1}{3}}} }$
b) ${\displaystyle m=-3(2)^{2}+4(2)-1=-3(4)+8-1=-12+7=-5}$

${\displaystyle 0=-5(2)+b\implies b=10}$
${\displaystyle \mathbf {y=-5x+10} }$
86. ${\displaystyle f(x)={\frac {2}{3}}x^{3}+{\frac {5}{2}}x^{2}+2x+1,\;\;\;(3,{\frac {95}{2}})}$
:${\displaystyle f'(x)=2x^{2}+5x+2}$

a) ${\displaystyle 2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf {x=-{\frac {1}{2}},-2} }$
/ b) ${\displaystyle m=2(3)^{2}+5(3)+2=18+15+2=35}$

${\displaystyle {\frac {95}{2}}=35(3)+b\implies b=-{\frac {115}{2}}}$
${\displaystyle \mathbf {y=35x-{\frac {115}{2}}} }$
:${\displaystyle f'(x)=2x^{2}+5x+2}$

a) ${\displaystyle 2x^{2}+5x+2=0=(2x+1)(x+2)\implies \mathbf {x=-{\frac {1}{2}},-2} }$
/ b) ${\displaystyle m=2(3)^{2}+5(3)+2=18+15+2=35}$

${\displaystyle {\frac {95}{2}}=35(3)+b\implies b=-{\frac {115}{2}}}$
${\displaystyle \mathbf {y=35x-{\frac {115}{2}}} }$
87. Find an equation of the tangent line to the graph defined by ${\displaystyle (x-y-1)^{3}=x\,}$ at the point (1,-1).
${\displaystyle 3(x-y-1)^{2}(1-y')=1}$

${\displaystyle 1-y'={\frac {1}{3(x-y-1)^{2}}}}$
${\displaystyle y'=1-{\frac {1}{3(x-y-1)^{2}}}}$
${\displaystyle m=1-{\frac {1}{3(1-(-1)-1)^{2}}}=1-{\frac {1}{3(1)^{2}}}={\frac {2}{3}}}$