Question

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

Answer 1

`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)`