How do you differentiate #f(x)=e^(cos(lnx))# using the chain rule?

1 Answer
Bill K.
Dec 7, 2015

The answer is #f'(x)=e^{cos(ln(x))} * -sin(ln(x)) * 1/x=-(sin(ln(x))e^{cos(ln(x))})/x#

Explanation:

Here are the details, based on using the formula #d/dx(f(g(x)))=f'(g(x)) * g'(x)# twice:

#f'(x)=e^{cos(ln(x))} * d/dx(cos(ln(x)))#

#=e^{cos(ln(x))} * -sin(ln(x)) * d/dx(ln(x))#

#=e^{cos(ln(x))} * -sin(ln(x)) * 1/x=-(sin(ln(x))e^{cos(ln(x))})/x#

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