Calculus/Differentiation/Basics of Differentiation/Exercises

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	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.

f(x)=x^{2}

2x

2x

2.

f(x)=2x+2

2

2

3.

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

x

x

4.

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

4x+4

4x+4

5.

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

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

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

6.

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

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

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

7.

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

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

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

8.

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

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

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

9.

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

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

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

Solutions

Prove the Constant Rule

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}}

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

Solutions

Find the Derivative by Rules

Find the derivative of the following functions:

Power Rule

11.

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

f'(x)=4x

f'(x)=4x

12.

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

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

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

13.

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

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

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

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}}}}

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}}}}

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

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}}}

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}}}}

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

Solutions

Product Rule

20.

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

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

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

21.

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

18x^{2}-6x+4

18x^{2}-6x+4

22.

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

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

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

23.

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

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

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

24.

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

30x^{2}+70x+6

30x^{2}+70x+6

25.

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

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

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)

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)

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)

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

Solutions

Quotient Rule

29.

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

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

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

30.

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

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

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

31.

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

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

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

32.

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

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

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

33.

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

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

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

34.

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

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

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

35.

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

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

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

36.

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

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

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

37.

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

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

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

38.

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

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

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

39.

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

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

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

40.

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

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

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

41.

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

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

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

42.

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

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

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

Solutions

Chain Rule

43.

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

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

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

44.

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

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

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

45.

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

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

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

46.

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

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

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

47.

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

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

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

48.

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

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

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

49.

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

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

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

50.

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

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

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

51.

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

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

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

52.

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

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

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

53.

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

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

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

Solutions

Exponentials

54.

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

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

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

55.

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

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

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

56.

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

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

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

57.

f(x)=4^{x}

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

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

Solutions

Logarithms

58.

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

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

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

59.

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

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

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

60.

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

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

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

61.

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

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

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

62.

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

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

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

Solutions

Trigonometric functions

63.

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

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

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

64.

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

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

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

Solutions

More Differentiation

65.

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

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

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

66.

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

3x^{2}+3

3x^{2}+3

67.

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

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

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

68.

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

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

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

69.

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

9x^{2}

9x^{2}

70.

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

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

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

71.

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

\ln(2)2^{x}

\ln(2)2^{x}

72.

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

2xe^{x^{2}}

2xe^{x^{2}}

73.

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

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

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

Solutions

Implicit Differentiation

Use implicit differentiation to find y'

74.

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

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

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

75.

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

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

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

Solutions

Logarithmic Differentiation

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

76.

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

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

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

77.

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

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

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

78.

y=(2x)^{2x}

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

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

79.

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

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

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

80.

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

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

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

Solutions

Equation of Tangent Line

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

81.

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

a)

x=0,-2

b)

y=15x-22

a)

x=0,-2

b)

y=15x-22

82.

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

a)

x=\pm 1

b)

y=-1

a)

x=\pm 1

b)

y=-1

83.

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

a)

x=2,-3

b)

y=-12x+6

a)

x=2,-3

b)

y=-12x+6

84.

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

a)

x=2^{-{\frac {4}{3}}}

b)

y={\frac {3x}{2}}+{\frac {3}{2}}

a)

x=2^{-{\frac {4}{3}}}

b)

y={\frac {3x}{2}}+{\frac {3}{2}}

85.

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

a)

x=1,{\frac {1}{3}}

b)

y=-5x+10

a)

x=1,{\frac {1}{3}}

b)

y=-5x+10

86.

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

a)

x=-{\frac {1}{2}},-2

/ b)

y=35x-{\frac {115}{2}}

a)

x=-{\frac {1}{2}},-2

/ b)

y=35x-{\frac {115}{2}}

87. 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}}

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

88. 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

y=-2x+2

Solutions

Higher Order Derivatives

89. What is the second derivative of

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

?

36x^{2}+6

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

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

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

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

Solutions

Advanced Understanding of Derivatives

91. Let

f^{\prime }(x)

be the derivative of

f(x)

. Prove the derivative of

-f(x)

is

-f^{\prime }(x)

.

Suppose $f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}$ . Let $g(x)=-f(x)$ .
${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$

Therefore, if

f^{\prime }(x)

is the derivative of

f(x)

, then

-f^{\prime }(x)

is the derivative of

-f(x)

.

\blacksquare

Suppose $f^{\prime }(x)=\lim _{h\to 0}{\dfrac {f(x+h)-f(h)}{h}}$ . Let $g(x)=-f(x)$ .
${\begin{aligned}g^{\prime }(x)&=\lim _{h\to 0}{\dfrac {g(x+h)-g(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-f(x+h)+f(h)}{h}}\\&=\lim _{h\to 0}{\dfrac {-\left(f(x+h)-f(h)\right)}{h}}\\&=-\lim _{h\to 0}{\dfrac {\left(f(x+h)-f(h)\right)}{h}}\\&=-f^{\prime }(x)\end{aligned}}$

Therefore, if

f^{\prime }(x)

is the derivative of

f(x)

, then

-f^{\prime }(x)

is the derivative of

-f(x)

.

\blacksquare

92. Suppose a continuous function

f(x)

has three roots on the interval of

-2\leq x\leq 2

. If

f(-2)=f(2)<0

, then what is ONE true guarantee of

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)

f(x)

is continuous. Ergo, the intermediate value theorem applies. There exists some

f(d)=c\in \left[f(-2),0\right]

such that

f(-2)<c<0

, where

d\in \left[-2,2\right]

.

(b) Rolle's Theorem does not apply for a non-differentiable function.

(c)

f(x)

is continuous. Ergo, the extreme value theorem applies. There exists a

-2\leq c\leq 2

so that

f(c)>f(x)

for all

x\in \left[-2,2\right]

.

These are examples only. More valid solutions may exist.

(a)

f(x)

is continuous. Ergo, the intermediate value theorem applies. There exists some

f(d)=c\in \left[f(-2),0\right]

such that

f(-2)<c<0

, where

d\in \left[-2,2\right]

.

(b) Rolle's Theorem does not apply for a non-differentiable function.

(c)

f(x)

is continuous. Ergo, the extreme value theorem applies. There exists a

-2\leq c\leq 2

so that

f(c)>f(x)

for all

x\in \left[-2,2\right]

.

93. Let

g(x)=f^{-1}(x)

, where

f^{-1}(x)

is the inverse of

f(x)

. Let

f(x)

be differentiable. What is

g^{\prime }(x)

? Else, why can

g^{\prime }(x)

not be determined?

g^{\prime }(x)={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}

.

g^{\prime }(x)={\frac {1}{f^{\prime }\left(f^{-1}(x)\right)}}

.

94. Let

f(x)={\begin{cases}2x^{2}+4,&x<-1\\ax^{3}-2x,&x\geq -1\end{cases}}

where

a

is a constant.

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

(a) The function

f(x)

is continuous but non-differentiable.

(b) The function

f(x)

is both continuous and differentiable.

(a)

a=-4

.

(b) There is no

a\in \mathbb {R}

that allows the following to be true. Proof in solutions.

(a)

a=-4

.

(b) There is no

a\in \mathbb {R}

that allows the following to be true. Proof in solutions.

Solutions

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