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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