Exam 4: More About Derivatives

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Let f(x)=tan(5x)f ( x ) = \tan \left( 5 ^ { x } \right) Then f(x)f ^ { \prime } ( x ) is

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Let f(x)=sin(cosx)f ( x ) = \sin ( \cos x ) Then f(x)f ^ { \prime } ( x ) is

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The linear approximation L(x)L ( x ) to f(x)=sinxf ( x ) = \sin x near x0=π6x _ { 0 } = - \frac { \pi } { 6 } is

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

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Let y=xx2y = x ^ { \ell ^ { x ^ { 2 } } } Then yy ^ { \prime } is

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Let f(x)=ecos(3x)f ( x ) = \mathrm { e } ^ { \cos ( 3 x ) } Then f(x)f ^ { \prime } ( x ) is

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Let y=sin(lnx)y = \sin ( \ln x ) Then yy ^ { \prime } is

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Let f(x)=sinh(ex)f ( x ) = \sinh \left( e ^ { x } \right) Then f(x)f ^ { \prime } ( x ) is

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Let f(x)=tanh1(ex)f ( x ) = \tanh ^ { - 1 } \left( e ^ { x } \right) Then f(x)f ^ { \prime } ( x ) is

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Using implicit differentiation on y=xsiny,yy = x \sin y , y ^ { \prime } is

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Let y=ln(tan1x)y = \ln \left( \tan ^ { - 1 } x \right) Then yy ^ { \prime } is

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Let y=2tanxy = 2 ^ { \tan x } Then dy is

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The linear approximation L(x)L ( x ) to f(x)=xf ( x ) = \sqrt { x } near x0=14x _ { 0 } = \frac { 1 } { 4 } is

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Let f(x)=tan1(3x)f ( x ) = \tan ^ { - 1 } \left( 3 ^ { x } \right) Then f(x)f ^ { \prime } ( x ) is

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Let y=sin(7x)y = \sin \left( 7 ^ { x } \right) Then dy is

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Letting f(x)=tanxf ( x ) = \tan x and x0=π4x _ { 0 } = \frac { \pi } { 4 } the approximation of tan 44° by differentials is

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Let y=xexy = x ^ {e ^ { x}} Then yy ^ { \prime } is

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Using implicit differentiation on 2x2xy+y2=4,y2 x ^ { 2 } - x y + y ^ { 2 } = 4 , y is

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Let f(x)=cosh1(2x)f ( x ) = \cosh ^ { - 1 } \left( 2 ^ { x } \right) Then f(x)f ^ { \prime } ( x ) is

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Let y=5(x2x+1)100y = 5 \left( x ^ { 2 } - x + 1 \right) ^ { 100 } Then yy ^ { \prime } is

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