# Calculus/Differentiation/Basics of Differentiation/Exercises

 ← Some Important Theorems Calculus L'Hôpital's rule → Differentiation/Basics of Differentiation/Exercises

## Find the Derivative by Definition

Find the derivative of the following functions using the limit definition of the derivative.

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

## Prove the Constant Rule

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

## Find the Derivative by Rules

Find the derivative of the following functions:

### Power Rule

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

### Product Rule

20. ${\displaystyle f(x)=(x^{4}+4x+2)(2x+3)}$
${\displaystyle 10x^{4}+12x^{3}+16x+16}$
${\displaystyle 10x^{4}+12x^{3}+16x+16}$
21. ${\displaystyle f(x)=(2x-1)(3x^{2}+2)}$
${\displaystyle 18x^{2}-6x+4}$
${\displaystyle 18x^{2}-6x+4}$
22. ${\displaystyle f(x)=(x^{3}-12x)(3x^{2}+2x)}$
${\displaystyle 15x^{4}+8x^{3}-108x^{2}-48x}$
${\displaystyle 15x^{4}+8x^{3}-108x^{2}-48x}$
23. ${\displaystyle f(x)=(2x^{5}-x)(3x+1)}$
${\displaystyle 36x^{5}+10x^{4}-6x-1}$
${\displaystyle 36x^{5}+10x^{4}-6x-1}$
24. ${\displaystyle f(x)=(5x^{2}+3)(2x+7)}$
${\displaystyle 30x^{2}+70x+6}$
${\displaystyle 30x^{2}+70x+6}$
25. ${\displaystyle f(x)=3x^{2}(5x^{2}+1)^{4}}$
${\displaystyle 6x(25x^{2}+1)(5x^{2}+1)^{3}}$
${\displaystyle 6x(25x^{2}+1)(5x^{2}+1)^{3}}$
26. ${\displaystyle f(x)=x^{3}(2x^{2}-x+4)^{4}}$
${\displaystyle x^{2}(2x^{2}-x+4)^{3}(22x^{2}-7x+12)}$
${\displaystyle x^{2}(2x^{2}-x+4)^{3}(22x^{2}-7x+12)}$
27. ${\displaystyle f(x)=5x^{2}(x^{3}-x+1)^{3}}$
${\displaystyle 5x(x^{3}-x+1)^{2}(11x^{3}-5x+2)}$
${\displaystyle 5x(x^{3}-x+1)^{2}(11x^{3}-5x+2)}$
28. ${\displaystyle f(x)=(2-x)^{6}(5+2x)^{4}}$
${\displaystyle 2(x-2)^{5}(2x+5)^{3}(10x+7)}$
${\displaystyle 2(x-2)^{5}(2x+5)^{3}(10x+7)}$

### Quotient Rule

29. ${\displaystyle f(x)={\frac {2x+1}{x+5}}}$
${\displaystyle f'(x)={\frac {9}{(x+5)^{2}}}}$
${\displaystyle f'(x)={\frac {9}{(x+5)^{2}}}}$
30. ${\displaystyle f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}}$
${\displaystyle f'(x)={\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}}$
${\displaystyle f'(x)={\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}}$
31. ${\displaystyle f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}}$
${\displaystyle f'(x)={\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}}$
${\displaystyle f'(x)={\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}}$
32. ${\displaystyle d(u)={\frac {u^{3}+2}{u^{3}}}}$
${\displaystyle d'(u)=-{\frac {6}{u^{4}}}}$
${\displaystyle d'(u)=-{\frac {6}{u^{4}}}}$
33. ${\displaystyle f(x)={\frac {x^{2}+x}{2x-1}}}$
${\displaystyle f'(x)={\frac {2x^{2}-2x-1}{(2x-1)^{2}}}}$
${\displaystyle f'(x)={\frac {2x^{2}-2x-1}{(2x-1)^{2}}}}$
34. ${\displaystyle f(x)={\frac {x+1}{2x^{2}+2x+3}}}$
${\displaystyle f'(x)={\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}}$
${\displaystyle f'(x)={\frac {-2x^{2}-4x+1}{(2x^{2}+2x+3)^{2}}}}$
35. ${\displaystyle f(x)={\frac {16x^{4}+2x^{2}}{x}}}$
${\displaystyle f'(x)=48x^{2}+2}$
${\displaystyle f'(x)=48x^{2}+2}$
36. ${\displaystyle f(x)={\frac {8x^{3}+2}{5x+5}}}$
${\displaystyle f'(x)={\frac {2(8x^{3}+12x^{2}-1)}{5(x+1)^{2}}}}$
${\displaystyle f'(x)={\frac {2(8x^{3}+12x^{2}-1)}{5(x+1)^{2}}}}$
37. ${\displaystyle f(x)={\frac {(3x-2)^{2}}{\sqrt {x}}}}$
${\displaystyle f'(x)={\frac {(3x-2)(9x+2)}{2x{\sqrt {x}}}}}$
${\displaystyle f'(x)={\frac {(3x-2)(9x+2)}{2x{\sqrt {x}}}}}$
38. ${\displaystyle f(x)={\frac {\sqrt {x}}{2x-1}}}$
${\displaystyle f'(x)={\frac {-(2x+1)}{2{\sqrt {x}}(2x-1)^{2}}}}$
${\displaystyle f'(x)={\frac {-(2x+1)}{2{\sqrt {x}}(2x-1)^{2}}}}$
39. ${\displaystyle f(x)={\frac {4x-3}{x+2}}}$
${\displaystyle f'(x)={\frac {11}{(x+2)^{2}}}}$
${\displaystyle f'(x)={\frac {11}{(x+2)^{2}}}}$
40. ${\displaystyle f(x)={\frac {4x+3}{2x-1}}}$
${\displaystyle f'(x)={\frac {-10}{(2x-1)^{2}}}}$
${\displaystyle f'(x)={\frac {-10}{(2x-1)^{2}}}}$
41. ${\displaystyle f(x)={\frac {x^{2}}{x+3}}}$
${\displaystyle f'(x)={\frac {x(x+6)}{(x+3)^{2}}}}$
${\displaystyle f'(x)={\frac {x(x+6)}{(x+3)^{2}}}}$
42. ${\displaystyle f(x)={\frac {x^{5}}{3-x}}}$
${\displaystyle f'(x)={\frac {x^{4}(-4x+15)}{(3-x)^{2}}}}$
${\displaystyle f'(x)={\frac {x^{4}(-4x+15)}{(3-x)^{2}}}}$

### Chain Rule

43. ${\displaystyle f(x)=(x+5)^{2}}$
${\displaystyle f'(x)=2(x+5)}$
${\displaystyle f'(x)=2(x+5)}$
44. ${\displaystyle g(x)=(x^{3}-2x+5)^{2}}$
${\displaystyle g'(x)=2(x^{3}-2x+5)(3x^{2}-2)}$
${\displaystyle g'(x)=2(x^{3}-2x+5)(3x^{2}-2)}$
45. ${\displaystyle f(x)={\sqrt {1-x^{2}}}}$
${\displaystyle f'(x)=-{\frac {x}{\sqrt {1-x^{2}}}}}$
${\displaystyle f'(x)=-{\frac {x}{\sqrt {1-x^{2}}}}}$
46. ${\displaystyle f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}}$
${\displaystyle f'(x)={\frac {6(4x^{3}+1)(2x+4)^{2}-(2x+4)^{3}(12x^{2})}{(4x^{3}+1)^{2}}}}$
${\displaystyle f'(x)={\frac {6(4x^{3}+1)(2x+4)^{2}-(2x+4)^{3}(12x^{2})}{(4x^{3}+1)^{2}}}}$
47. ${\displaystyle f(x)=(2x+1){\sqrt {2x+2}}}$
${\displaystyle f'(x)=2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}}$
${\displaystyle f'(x)=2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}}$
48. ${\displaystyle f(x)={\frac {2x+1}{\sqrt {2x+2}}}}$
${\displaystyle f'(x)={\frac {2x+3}{(2x+2){\sqrt {2x+2}}}}}$
${\displaystyle f'(x)={\frac {2x+3}{(2x+2){\sqrt {2x+2}}}}}$
49. ${\displaystyle f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}}$
${\displaystyle f'(x)={\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)}$
${\displaystyle f'(x)={\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)}$
50. ${\displaystyle f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}}$
${\displaystyle f'(x)={\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}}$
${\displaystyle f'(x)={\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}}$
51. ${\displaystyle f(x)={\sqrt {x^{3}+1}}(x^{2}-1)}$
${\displaystyle f'(x)={\frac {3x^{2}(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}}$
${\displaystyle f'(x)={\frac {3x^{2}(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}}$
52. ${\displaystyle f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}}$
${\displaystyle f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}$
${\displaystyle f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)}$
53. ${\displaystyle f(x)={\sqrt {1+x^{2}}}}$
${\displaystyle f'(x)={\frac {x}{\sqrt {1+x^{2}}}}}$
${\displaystyle f'(x)={\frac {x}{\sqrt {1+x^{2}}}}}$

### Exponentials

54. ${\displaystyle f(x)=(3x^{2}+e)e^{2x}}$
${\displaystyle f'(x)=6xe^{2x}+2e^{2x}(3x^{2}+e)}$
${\displaystyle f'(x)=6xe^{2x}+2e^{2x}(3x^{2}+e)}$
55. ${\displaystyle f(x)=e^{2x^{2}+3x}}$
${\displaystyle f'(x)=(4x+3)e^{2x^{2}+3x}}$
${\displaystyle f'(x)=(4x+3)e^{2x^{2}+3x}}$
56. ${\displaystyle f(x)=e^{e^{2x^{2}+1}}}$
${\displaystyle f'(x)=4xe^{2x^{2}+1+e^{2x^{2}+1}}}$
${\displaystyle f'(x)=4xe^{2x^{2}+1+e^{2x^{2}+1}}}$
57. ${\displaystyle f(x)=4^{x}}$
${\displaystyle f'(x)=\ln(4)4^{x}}$
${\displaystyle f'(x)=\ln(4)4^{x}}$

### Logarithms

58. ${\displaystyle f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln(x)}$
${\displaystyle f'(x)=3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-3}}{2{\sqrt {x^{3}-2}}}}+{\frac {1}{x}}}$
${\displaystyle f'(x)=3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-3}}{2{\sqrt {x^{3}-2}}}}+{\frac {1}{x}}}$
59. ${\displaystyle f(x)=\ln(x)-2e^{x}+{\sqrt {x}}}$
${\displaystyle f'(x)={\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}}$
${\displaystyle f'(x)={\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}}$
60. ${\displaystyle f(x)=\ln(\ln(x^{3}(x+1)))}$
${\displaystyle f'(x)={\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}}$
${\displaystyle f'(x)={\frac {4x^{3}+3x^{2}}{x^{3}(x+1)\ln(x^{3}(x+1))}}}$
61. ${\displaystyle f(x)=\ln(2x^{2}+3x)}$
${\displaystyle f'(x)={\frac {4x+3}{2x^{2}+3x}}}$
${\displaystyle f'(x)={\frac {4x+3}{2x^{2}+3x}}}$
62. ${\displaystyle f(x)=\log _{4}(x)+2\ln(x)}$
${\displaystyle f'(x)={\frac {1}{x\ln(4)}}+{\frac {2}{x}}}$
${\displaystyle f'(x)={\frac {1}{x\ln(4)}}+{\frac {2}{x}}}$

### Trigonometric functions

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

## More Differentiation

65. ${\displaystyle {\frac {d}{dx}}[(x^{3}+5)^{10}]}$
${\displaystyle 30x^{2}(x^{3}+5)^{9}}$
${\displaystyle 30x^{2}(x^{3}+5)^{9}}$
66. ${\displaystyle {\frac {d}{dx}}[x^{3}+3x]}$
${\displaystyle 3x^{2}+3}$
${\displaystyle 3x^{2}+3}$
67. ${\displaystyle {\frac {d}{dx}}[(x+4)(x+2)(x-3)]}$
${\displaystyle (x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}$
${\displaystyle (x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)}$
68. ${\displaystyle {\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]}$
${\displaystyle -{\frac {x+2}{3x^{3}}}}$
${\displaystyle -{\frac {x+2}{3x^{3}}}}$
69. ${\displaystyle {\frac {d}{dx}}[3x^{3}]}$
${\displaystyle 9x^{2}}$
${\displaystyle 9x^{2}}$
70. ${\displaystyle {\frac {d}{dx}}[x^{4}\sin(x)]}$
${\displaystyle 4x^{3}\sin(x)+x^{4}\cos(x)}$
${\displaystyle 4x^{3}\sin(x)+x^{4}\cos(x)}$
71. ${\displaystyle {\frac {d}{dx}}[2^{x}]}$
${\displaystyle \ln(2)2^{x}}$
${\displaystyle \ln(2)2^{x}}$
72. ${\displaystyle {\frac {d}{dx}}[e^{x^{2}}]}$
${\displaystyle 2xe^{x^{2}}}$
${\displaystyle 2xe^{x^{2}}}$
73. ${\displaystyle {\frac {d}{dx}}[e^{2^{x}}]}$
${\displaystyle \ln(2)2^{x}e^{2^{x}}}$
${\displaystyle \ln(2)2^{x}e^{2^{x}}}$

## Implicit Differentiation

Use implicit differentiation to find y'

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

## Logarithmic Differentiation

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

76. ${\displaystyle y=x({\sqrt[{4}]{1-x^{3}}})}$
${\displaystyle y'={\sqrt[{4}]{1-x^{3}}}+{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}}$
${\displaystyle y'={\sqrt[{4}]{1-x^{3}}}+{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}}$
77. ${\displaystyle y={\sqrt {x+1 \over 1-x}}\,}$
${\displaystyle y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}\left({\frac {1}{x+1}}+{\frac {1}{1-x}}\right)}$
${\displaystyle y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}\left({\frac {1}{x+1}}+{\frac {1}{1-x}}\right)}$
78. ${\displaystyle y=(2x)^{2x}}$
${\displaystyle y'=(2x)^{2x}(2\ln(2x)+2)}$
${\displaystyle y'=(2x)^{2x}(2\ln(2x)+2)}$
79. ${\displaystyle y=(x^{3}+4x)^{3x+1}}$
${\displaystyle y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})}$
${\displaystyle y'=(x^{3}+4x)^{3x+1}(3\ln(x^{3}+4x)+{\frac {(3x+1)(3x^{2}+4)}{x^{3}+4x}})}$
80. ${\displaystyle y=(6x)^{\cos(x)+1}}$
${\displaystyle y'=6x^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})}$
${\displaystyle y'=6x^{\cos(x)+1}(-\sin(x)\ln(6x)+{\frac {\cos(x)+1}{x}})}$

## 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. ${\displaystyle f(x)={\frac {x^{3}}{3}}+x^{2}+5,\qquad (3,23)}$
a) ${\displaystyle x=0,-2}$
b) ${\displaystyle y=15x-22}$
a) ${\displaystyle x=0,-2}$
b) ${\displaystyle y=15x-22}$
82. ${\displaystyle f(x)=x^{3}-3x+1,\qquad (1,-1)}$
a) ${\displaystyle x=\pm 1}$
b) ${\displaystyle y=-1}$
a) ${\displaystyle x=\pm 1}$
b) ${\displaystyle y=-1}$
83. ${\displaystyle f(x)={\frac {2x^{3}}{3}}+x^{2}-12x+6,\qquad (0,6)}$
a) ${\displaystyle x=2,-3}$
b) ${\displaystyle y=-12x+6}$
a) ${\displaystyle x=2,-3}$
b) ${\displaystyle y=-12x+6}$
84. ${\displaystyle f(x)=2x+{\frac {1}{\sqrt {x}}},\qquad (1,3)}$
a) ${\displaystyle x=2^{-{\frac {4}{3}}}}$
b) ${\displaystyle y={\frac {3x}{2}}+{\frac {3}{2}}}$
a) ${\displaystyle x=2^{-{\frac {4}{3}}}}$
b) ${\displaystyle y={\frac {3x}{2}}+{\frac {3}{2}}}$
85. ${\displaystyle f(x)=(x^{2}+1)(2-x),\qquad (2,0)}$
a) ${\displaystyle x=1,{\frac {1}{3}}}$
b) ${\displaystyle y=-5x+10}$
a) ${\displaystyle x=1,{\frac {1}{3}}}$
b) ${\displaystyle y=-5x+10}$
86. ${\displaystyle f(x)={\frac {2x^{3}}{3}}+{\frac {5x^{2}}{2}}+2x+1,\qquad \left(3,{\frac {95}{2}}\right)}$
a) ${\displaystyle x=-{\frac {1}{2}},-2}$
/ b) ${\displaystyle y=35x-{\frac {115}{2}}}$
a) ${\displaystyle x=-{\frac {1}{2}},-2}$
/ b) ${\displaystyle y=35x-{\frac {115}{2}}}$
87. Find an equation of the tangent line to the graph defined by ${\displaystyle (x-y-1)^{3}=x}$  at the point (1,-1).
${\displaystyle y={\frac {2x}{3}}-{\frac {5}{3}}}$
${\displaystyle y={\frac {2x}{3}}-{\frac {5}{3}}}$
88. Find an equation of the tangent line to the graph defined by ${\displaystyle e^{xy}+x^{2}=y^{2}}$  at the point (1,0).
${\displaystyle y=-2x+2}$
${\displaystyle y=-2x+2}$

## Higher Order Derivatives

89. What is the second derivative of ${\displaystyle 3x^{4}+3x^{2}+2x}$ ?
${\displaystyle 36x^{2}+6}$
${\displaystyle 36x^{2}+6}$
90. Use induction to prove that the (n+1)th derivative of a n-th order polynomial is 0.

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

${\displaystyle {\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0}$

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

${\displaystyle {\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0}$

91. Let ${\displaystyle f^{\prime }(x)}$  be the derivative of ${\displaystyle f(x)}$ . Prove the derivative of ${\displaystyle -f(x)}$  is ${\displaystyle -f^{\prime }(x)}$ .

Suppose ${\displaystyle f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}}$ . Let ${\displaystyle g(x)=-f(x)}$ .
{\displaystyle {\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}}}

Therefore, if ${\displaystyle f^{\prime }(x)}$  is the derivative of ${\displaystyle f(x)}$ , then ${\displaystyle -f^{\prime }(x)}$  is the derivative of ${\displaystyle -f(x)}$ . ${\displaystyle \blacksquare }$

Suppose ${\displaystyle f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}}$ . Let ${\displaystyle g(x)=-f(x)}$ .
{\displaystyle {\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}}}

Therefore, if ${\displaystyle f^{\prime }(x)}$  is the derivative of ${\displaystyle f(x)}$ , then ${\displaystyle -f^{\prime }(x)}$  is the derivative of ${\displaystyle -f(x)}$ . ${\displaystyle \blacksquare }$
92. Suppose a continuous function ${\displaystyle f(x)}$  has three roots on the interval of ${\displaystyle -2\leq x\leq 2}$ . If ${\displaystyle f(-2)=f(2)<0}$ , then what is ONE true guarantee of ${\displaystyle f(x)}$  using
(a) the Intermediate Value Theorem;
(b) Rolle's Theorem;
(c) the Extreme Value Theorem.
These are examples only. More valid solutions may exist.
(a) ${\displaystyle f(x)}$  is continuous. Ergo, the intermediate value theorem applies. There exists some ${\displaystyle f(d)=c\in \left[f(-2),0\right]}$  such that ${\displaystyle f(-2) , where ${\displaystyle d\in \left[-2,2\right]}$ .
(b) Rolle's Theorem does not apply for a non-differentiable function.
(c) ${\displaystyle f(x)}$  is continuous. Ergo, the extreme value theorem applies. There exists a ${\displaystyle -2\leq c\leq 2}$  so that ${\displaystyle f(c)>f(x)}$  for all ${\displaystyle x\in \left[-2,2\right]}$ .
These are examples only. More valid solutions may exist.
(a) ${\displaystyle f(x)}$  is continuous. Ergo, the intermediate value theorem applies. There exists some ${\displaystyle f(d)=c\in \left[f(-2),0\right]}$  such that ${\displaystyle f(-2) , where ${\displaystyle d\in \left[-2,2\right]}$ .
(b) Rolle's Theorem does not apply for a non-differentiable function.
(c) ${\displaystyle f(x)}$  is continuous. Ergo, the extreme value theorem applies. There exists a ${\displaystyle -2\leq c\leq 2}$  so that ${\displaystyle f(c)>f(x)}$  for all ${\displaystyle x\in \left[-2,2\right]}$ .
93. Let ${\displaystyle g(x)=f^{-1}(x)}$ , where ${\displaystyle f^{-1}(x)}$  is the inverse of ${\displaystyle f(x)}$ . Let ${\displaystyle f(x)}$  be differentiable. What is ${\displaystyle g^{\prime }(x)}$ ? Else, why can ${\displaystyle g^{\prime }(x)}$  not be determined?
${\displaystyle g^{\prime }(x)={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}}$ .
${\displaystyle g^{\prime }(x)={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}}$ .
94. Let ${\displaystyle f(x)={\begin{cases}2x^{2}+4,&x<-1\\ax^{3}-2x,&x\geq -1\end{cases}}}$  where ${\displaystyle a}$  is a constant.

Find a value, if possible, for ${\displaystyle a}$  that allows each of the following to be true. If not possible, prove that it cannot be done.

(a) The function ${\displaystyle f(x)}$  is continuous but non-differentiable.
(b) The function ${\displaystyle f(x)}$  is both continuous and differentiable.
(a) ${\displaystyle a=-4}$ .
(b) There is no ${\displaystyle a\in \mathbb {R} }$  that allows the following to be true. Proof in solutions.
(a) ${\displaystyle a=-4}$ .
(b) There is no ${\displaystyle a\in \mathbb {R} }$  that allows the following to be true. Proof in solutions.
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