Exam 4: More About Derivatives

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Let f(x)=cosh(3x)f ( x ) = \cosh \left( 3 ^ { x } \right) Then f(x)f ^ { \prime } ( x ) is

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Letting f(x)=xf ( x ) = \sqrt { x } and x0=25x _ { 0 } = 25 the approximation of 24\sqrt { 24 } by differentials is

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Letting f(x)=xf ( x ) = \sqrt { x } and x0=25,x _ { 0 } = 25 , the approximation of 25.1\sqrt { 25.1 } by differentials is

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Let xy=ex ^ { y } = e \text {. } Then yy ^ { \prime } is

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Let f(x)=tan3(5x)f ( x ) = \tan ^ { 3 } ( 5 x ) Then f(x)f ^ { \prime } ( x ) is

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Let y=tan3xy = \tan ^ { 3 } x Then yy ^ { \prime } is

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Let f(x)=sinh(sinx)f ( x ) = \sinh ( \sin x ) Then f(x)f ^ { \prime } ( x ) is

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Let f(x)=tan1(x)f ( x ) = \tan ^ { - 1 } ( \sqrt { x } ) Then f(x)f ^ { \prime } ( x ) is

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Let f(x)=cosh(x2+1)f ( x ) = \cosh \left( x ^ { 2 } + 1 \right) Then f(x)f ^ { \prime } ( x ) is

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Let f(x)=sinh1(sinx)f ( x ) = \sinh ^ { - 1 } ( \sin x ) Then f(x)f ^ { \prime } ( x ) is

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Let f(x)=2xcoshxf ( x ) = 2 ^ { x } \cosh x Then f(x)f ^ { \prime } ( x ) is

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Let y=log2(tan3x)y = \log _ { 2 } \left( \tan ^ { 3 } x \right) Then dy is

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Let f(x)=x2sin1xf ( x ) = x ^ { 2 } \sin ^ { - 1 } x Then f(x)f ^ { \prime } ( x ) is

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Let h=fg.h = f \circ g . Then g(2)=3g ( 2 ) = 3 , g(2)=1g ^ { \prime } ( 2 ) = - 1 and f(3)=6f ^ { \prime } ( 3 ) = 6 Then h(2)h ^ { \prime } ( 2 ) is

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Using implicit differentiation on y=(x+2y)2,yy = ( x + 2 y ) ^ { 2 } , y ^ { \prime } is

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Let y=ln(xy)y = \ln ( x y ) Using implicit differentiation, yy ^ { \prime } is

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Let f(x)=coth(3x)f ( x ) = \operatorname { coth } ( 3 x ) Then f(x)f ^ { \prime } ( x ) is

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Let y=sinu+cosuy = \sin u + \cos u and u=4x2+5u = 4 x ^ { 2 } + 5 By the Chain Rule, dydx\frac { d y } { d x } is

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Let f(x)=ln(log2x)f ( x ) = \ln \left( \log _ { 2 } x \right) Then f(x)f ^ { \prime } ( x ) is

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Let y=ln(5x)y = \ln \left( 5 ^ { x } \right) Then dy is

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