Question

Question #56c3f

Answer 1

`(dy)/(dx)=-2cos(x^2+1)e^(-2x)(2xsin(x^2+1)+cos(x^2+1))`

Explanation:

`y=e^(-2x)cos^2(x^2+1)=uv`

If `y=f(x)g(x) => (dy)/(dx)=f(x)g'(x)+f'(x)g(x)`

`f(x)=e^(-2x)`
`f'(x)=-2e^(-2x)`

`g(x)=cos^2(x^2+1)=(cos(x^2+1))^2`
`g'(x)=2cos(x^2+1) * d/dx[cos(x^2+1)]`
`g'(x)=2cos(x^2+1) * -2xsin(x^2+1)`
`g'(x)=-4xsin(x^2+1)cos(x^2+1)`

`(dy)/(dx)=-4xsin(x^2+1)cos(x^2+1)e^(-2x)-2e^(-2x)cos^2(x^2+1)=-2cos(x^2+1)e^(-2x)(2xsin(x^2+1)+cos(x^2+1))`