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

1 Answer
Shiva Prakash M V
Mar 2, 2018

#dy/dx=2xe^(x^2cosx)(xsinx-2cosx)#

Explanation:

#f(x)=-2e^(x^2cosx)#
Let
#y=f(x)#
Then
#y=-2e^(x^2cosx)#
Let
#u=e^(x^2cosx)#
Taking logarithms
#lnu=x^2cosx#
Differentiating wrt x
#1/u(du)/dx=x^2(-sinx)+cosx(2x)#
#1/u(du)/dx=-x^2sinx+2xcosx#
#(du)/dx=u(-x^2sinx+2xcosx)#
#y=-2e^(x^2cosx)#
#y=-2u#
Differentiating
#dy/dx=-2(du)/dx#
#(du)/dx=u(-x^2sinx+2xcosx)#
#dy/dx=-2(u(-x^2sinx+2xcosx))#
#dy/dx=2u(x^2sinx-2xcosx)#
#u=e^(x^2cosx)#
#dy/dx=2e^(x^2cosx)(x^2sinx-2xcosx)#
#dy/dx=2xe^(x^2cosx)(xsinx-2cosx)#

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