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How do you differentiate #f(x)=sin(e^(3-x)) # using the chain rule?

1 Answer

Apr 21, 2018

#f'(x)=-e^(3-x)cos(e^(3-x))#

Explanation:

The Chain Rule, when applied to the sine, tells us that

#d/dxsin(u)=cosu*(du)/dx,# where #u# is some function in terms of #x.#

Here, #u=e^(3-x)#, so

#f'(x)=cos(e^(3-x))*d/dxe^(3-x)#

#d/dxe^(3-x)=e^(3-x)*d/dx(3-x)=-e^(3-x)# so we get

#f'(x)=-e^(3-x)cos(e^(3-x))#

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