Calculus/Differentiation/Basics of Differentiation/Exercises

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	Differentiation/Basics of Differentiation/Exercises

Find the Derivative by DefinitionEdit

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

1.

f(x)=x^{2}

$2x$

2.

f(x)=2x+2

$2$

3.

f(x)={\frac {x^{2}}{2}}

$x$

4.

f(x)=2x^{2}+4x+4

$4x+4$

5.

f(x)={\sqrt {x+2}}

${\frac {1}{2{\sqrt {x+2}}}}$

6.

f(x)={\frac {1}{x}}

$-{\frac {1}{x^{2}}}$

7.

f(x)={\frac {3}{x+1}}

${\frac {-3}{(x+1)^{2}}}$

8.

f(x)={\frac {1}{\sqrt {x+1}}}

${\frac {-1}{2(x+1)^{3/2}}}$

9.

f(x)={\frac {x}{x+2}}

${\frac {2}{(x+2)^{2}}}$

Solutions

Prove the Constant RuleEdit

10. Use the definition of the derivative to prove that for any fixed real number

c

,

{\frac {d}{dx}}[c\cdot f(x)]=c\cdot {\frac {d}{dx}}[f(x)]

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

Solutions

Find the Derivative by RulesEdit

Find the derivative of the following functions:

Power RuleEdit

11.

f(x)=2x^{2}+4

$f'(x)=4x$

12.

f(x)=3{\sqrt[{3}]{x}}\,

$f'(x)={\frac {1}{\sqrt[{3}]{x^{2}}}}$

13.

f(x)=2x^{5}+8x^{2}+x-78

$f'(x)=10x^{4}+16x+1$

14.

f(x)=7x^{7}+8x^{5}+x^{3}+x^{2}-x

$f'(x)=49x^{6}+40x^{4}+3x^{2}+2x-1$

15.

f(x)={\frac {1}{x^{2}}}+3x^{\frac {1}{3}}

$f'(x)={\frac {-2}{x^{3}}}+{\frac {1}{\sqrt[{3}]{x^{2}}}}$

16.

f(x)=3x^{15}+{\frac {x^{2}}{17}}+{\frac {2}{\sqrt {x}}}

$f'(x)=45x^{14}+{\frac {2x}{17}}-{\frac {1}{x{\sqrt {x}}}}$

17.

f(x)={\frac {3}{x^{4}}}-{\sqrt[{4}]{x}}+x

$f'(x)={\frac {-12}{x^{5}}}-{\frac {1}{4{\sqrt[{4}]{x^{3}}}}}+1$

18.

f(x)=6x^{1/3}-x^{0.4}+{\frac {9}{x^{2}}}

$f'(x)={\frac {2}{\sqrt[{3}]{x^{2}}}}-{\frac {0.4}{x^{0.6}}}-{\frac {18}{x^{3}}}$

19.

f(x)={\frac {1}{\sqrt[{3}]{x}}}+{\sqrt {x}}

$f'(x)={\frac {-1}{3{\sqrt[{3}]{x^{4}}}}}+{\frac {1}{2{\sqrt {x}}}}$

Solutions

Product RuleEdit

20.

f(x)=(x^{4}+4x+2)(2x+3)

$10x^{4}+12x^{3}+16x+16$

21.

f(x)=(2x-1)(3x^{2}+2)

$18x^{2}-6x+4$

22.

f(x)=(x^{3}-12x)(3x^{2}+2x)

$15x^{4}+8x^{3}-108x^{2}-48x$

23.

f(x)=(2x^{5}-x)(3x+1)

$36x^{5}+10x^{4}-6x-1$

24.

f(x)=(5x^{2}+3)(2x+7)

$30x^{2}+70x+6$

25.

f(x)=3x^{2}(5x^{2}+1)^{4}

$6x(25x^{2}+1)(5x^{2}+1)^{3}$

26.

f(x)=x^{3}(2x^{2}-x+4)^{4}

$x^{2}(2x^{2}-x+4)^{3}(22x^{2}-7x+12)$

27.

f(x)=5x^{2}(x^{3}-x+1)^{3}

$5x(x^{3}-x+1)^{2}(11x^{3}-5x+2)$

28.

f(x)=(2-x)^{6}(5+2x)^{4}

$2(x-2)^{5}(2x+5)^{3}(10x+7)$

Solutions

Quotient RuleEdit

24.

f(x)={\frac {2x+1}{x+5}}

$f'(x)={\frac {9}{(x+5)^{2}}}$

25.

f(x)={\frac {3x^{4}+2x+2}{3x^{2}+1}}

$f'(x)={\frac {18x^{5}+12x^{3}-6x^{2}-12x+2}{(3x^{2}+1)^{2}}}$

26.

f(x)={\frac {x^{\frac {3}{2}}+1}{x+2}}

$f'(x)={\frac {x{\sqrt {x}}+6{\sqrt {x}}-2}{2(x+2)^{2}}}$

27.

d(u)={\frac {u^{3}+2}{u^{3}}}

$d'(u)=-{\frac {6}{u^{4}}}$

28.

f(x)={\frac {x^{2}+x}{2x-1}}

$f'(x)={\frac {2x^{2}-2x-1}{(2x-1)^{2}}}$

29.

f(x)={\frac {x+1}{2x^{2}+2x+3}}

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

30.

f(x)={\frac {16x^{4}+2x^{2}}{x}}

$f'(x)=48x^{2}+2$

f(x)={\frac {8x^{3}+2}{5x+5}}

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

f(x)={\frac {(3x-2)^{2}}{\sqrt {x}}}

$f'(x)={\frac {(3x-2)(9x+2)}{2x{\sqrt {x}}}}$

f(x)={\frac {\sqrt {x}}{2x-1}}

$f'(x)={\frac {-(2x+1)}{2{\sqrt {x}}(2x-1)^{2}}}$

f(x)={\frac {4x-3}{x+2}}

$f'(x)={\frac {11}{(x+2)^{2}}}$

f(x)={\frac {4x+3}{2x-1}}

$f'(x)={\frac {-10}{(2x-1)^{2}}}$

f(x)={\frac {x^{2}}{x+3}}

$f'(x)={\frac {x(x+6)}{(x+3)^{2}}}$

f(x)={\frac {x^{5}}{3-x}}

$f'(x)={\frac {x^{4}(-4x+15)}{(3-x)^{2}}}$

Solutions

Chain RuleEdit

31.

f(x)=(x+5)^{2}

$f'(x)=2(x+5)$

32.

g(x)=(x^{3}-2x+5)^{2}

$g'(x)=2(x^{3}-2x+5)(3x^{2}-2)$

33.

f(x)={\sqrt {1-x^{2}}}

$f'(x)=-{\frac {x}{\sqrt {1-x^{2}}}}$

34.

f(x)={\frac {(2x+4)^{3}}{4x^{3}+1}}

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

35.

f(x)=(2x+1){\sqrt {2x+2}}

$f'(x)=2{\sqrt {2x+2}}+{\frac {2x+1}{\sqrt {2x+2}}}$

36.

f(x)={\frac {2x+1}{\sqrt {2x+2}}}

$f'(x)={\frac {2x+3}{(2x+2){\sqrt {2x+2}}}}$

37.

f(x)={\sqrt {2x^{2}+1}}(3x^{4}+2x)^{2}

$f'(x)={\frac {2x(3x^{4}+2x)^{2}}{\sqrt {2x^{2}+1}}}+{\sqrt {2x^{2}+1}}(2)(3x^{4}+2x)(12x^{3}+2)$

38.

f(x)={\frac {2x+3}{(x^{4}+4x+2)^{2}}}

$f'(x)={\frac {2(x^{4}+4x+2)^{2}-2(2x+3)(x^{4}+4x+2)(4x^{3}+4)}{(x^{4}+4x+2)^{4}}}$

39.

f(x)={\sqrt {x^{3}+1}}(x^{2}-1)

$f'(x)={\frac {3x^{2}(x^{2}-1)}{2{\sqrt {x^{3}+1}}}}+2x{\sqrt {x^{3}+1}}$

40.

f(x)=((2x+3)^{4}+4(2x+3)+2)^{2}

$f'(x)=2((2x+3)^{4}+4(2x+3)+2)(8(2x+3)^{3}+8)$

41.

f(x)={\sqrt {1+x^{2}}}

$f'(x)={\frac {x}{\sqrt {1+x^{2}}}}$

Solutions

ExponentialsEdit

42.

f(x)=(3x^{2}+e)e^{2x}

$f'(x)=6xe^{2x}+2e^{2x}(3x^{2}+e)$

43.

f(x)=e^{2x^{2}+3x}

$f'(x)=(4x+3)e^{2x^{2}+3x}$

44.

f(x)=e^{e^{2x^{2}+1}}

$f'(x)=4xe^{2x^{2}+1+e^{2x^{2}+1}}$

45.

f(x)=4^{x}

$f'(x)=\ln(4)4^{x}$

Solutions

LogarithmsEdit

46.

f(x)=2^{x-3}\cdot 3{\sqrt {x^{3}-2}}+\ln(x)

$f'(x)=3\ln(2)2^{x-3}{\sqrt {x^{3}-2}}+{\frac {9x^{2}2^{x-3}}{2{\sqrt {x^{3}-2}}}}+{\frac {1}{x}}$

47.

f(x)=\ln(x)-2e^{x}+{\sqrt {x}}

$f'(x)={\frac {1}{x}}-2e^{x}+{\frac {1}{2{\sqrt {x}}}}$

48.

f(x)=\ln(\ln(x^{3}(x+1)))

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

49.

f(x)=\ln(2x^{2}+3x)

$f'(x)={\frac {4x+3}{2x^{2}+3x}}$

50.

f(x)=\log _{4}(x)+2\ln(x)

$f'(x)={\frac {1}{x\ln(4)}}+{\frac {2}{x}}$

Solutions

Trigonometric functionsEdit

51.

f(x)=3e^{x}-4\cos(x)-{\frac {\ln(x)}{4}}

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

52.

f(x)=\sin(x)+\cos(x)

$f'(x)=\cos(x)-\sin(x)$

Solutions

More DifferentiationEdit

53.

{\frac {d}{dx}}[(x^{3}+5)^{10}]

$30x^{2}(x^{3}+5)^{9}$

54.

{\frac {d}{dx}}[x^{3}+3x]

$3x^{2}+3$

55.

{\frac {d}{dx}}[(x+4)(x+2)(x-3)]

$(x+2)(x-3)+(x+4)(x-3)+(x+4)(x+2)$

56.

{\frac {d}{dx}}[{\frac {x+1}{3x^{2}}}]

$-{\frac {x+2}{3x^{3}}}$

57.

{\frac {d}{dx}}[3x^{3}]

$9x^{2}$

58.

{\frac {d}{dx}}[x^{4}\sin(x)]

$4x^{3}\sin(x)+x^{4}\cos(x)$

59.

{\frac {d}{dx}}[2^{x}]

$\ln(2)2^{x}$

60.

{\frac {d}{dx}}[e^{x^{2}}]

$2xe^{x^{2}}$

61.

{\frac {d}{dx}}[e^{2^{x}}]

$\ln(2)2^{x}e^{2^{x}}$

Solutions

Implicit DifferentiationEdit

Use implicit differentiation to find y'

62.

x^{3}+y^{3}=xy

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

63.

(2x+y)^{4}+3x^{2}+3y^{2}={\frac {x}{y}}+1

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

Solutions

Logarithmic DifferentiationEdit

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

64.

y=x({\sqrt[{4}]{1-x^{3}}})

$y'={\sqrt[{4}]{1-x^{3}}}+{\frac {3x^{3}}{4(1-x^{3})^{\frac {3}{4}}}}$

65.

y={\sqrt {x+1 \over 1-x}}\,

$y'={\frac {1}{2}}{\sqrt {\frac {x+1}{1-x}}}\left({\frac {1}{x+1}}+{\frac {1}{1-x}}\right)$

66.

y=(2x)^{2x}

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

67.

y=(x^{3}+4x)^{3x+1}

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

68.

y=(6x)^{\cos(x)+1}

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

Solutions

Equation of Tangent LineEdit

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.

69.

f(x)={\frac {x^{3}}{3}}+x^{2}+5,\qquad (3,23)

a) $x=0,-2$
b) $y=15x-22$

70.

f(x)=x^{3}-3x+1,\qquad (1,-1)

a) $x=\pm 1$
b) $y=-1$

71.

f(x)={\frac {2x^{3}}{3}}+x^{2}-12x+6,\qquad (0,6)

a) $x=2,-3$
b) $y=-12x+6$

72.

f(x)=2x+{\frac {1}{\sqrt {x}}},\qquad (1,3)

a) $x=2^{-{\frac {4}{3}}}$
b) $y={\frac {3x}{2}}+{\frac {3}{2}}$

73.

f(x)=(x^{2}+1)(2-x),\qquad (2,0)

a) $x=1,{\frac {1}{3}}$
b) $y=-5x+10$

74.

f(x)={\frac {2x^{3}}{3}}+{\frac {5x^{2}}{2}}+2x+1,\qquad (3,{\frac {95}{2}})

a) $x=-{\frac {1}{2}},-2$
/ b) $y=35x-{\frac {115}{2}}$

75. Find an equation of the tangent line to the graph defined by

(x-y-1)^{3}=x

at the point (1,-1).

$y={\frac {2x}{3}}-{\frac {5}{3}}$

76. Find an equation of the tangent line to the graph defined by

e^{xy}+x^{2}=y^{2}

at the point (1,0).

$y=-2x+2$

Solutions

Higher Order DerivativesEdit

77. What is the second derivative of

3x^{4}+3x^{2}+2x

?

$36x^{2}+6$

78. 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, $c$ . ${\frac {dc}{dx}}=0$
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.
${\frac {d^{n+1}}{dx^{n+1}}}f(x)={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n}+P(x))={\frac {d^{n+1}}{dx^{n+1}}}(cx^{n})+{\frac {d^{n+1}}{dx^{n+1}}}P(x)={\frac {d^{n}}{dx^{n}}}(cnx^{n-1})+{\frac {d}{dx}}{\frac {d^{n}}{dx^{n}}}P(x)=0+{\frac {d}{dx}}0=0$