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How do you differentiate # f(x)=sin(e^((lnx-2)^2 ))# using the chain rule.?

1 Answer

Apr 14, 2016

#f'(x)=cos(e^((lnx-2)^2)) xxe^((lnx-2)^2) xx 2(lnx-2) xx 1/x#

Explanation:

#f(x)=sin x, g(x)= e^x, h(x)=x^2, r(x)=lnx-2#

#f(g(h(r(x))))=f'(g(h(r(x))))xxg'(h(r(x)))^##h'(r(x))xxr'(x)#

#f'(x)=cos(e^((lnx-2)^2)) xxe^((lnx-2)^2) xx 2(lnx-2) xx 1/x#

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