If `y=e^xsinx,` find `dy/dx`

Answer

`e^(sin x^(2))cos x^(2)2x`

Explanation

`"Let "y=e^(sinx^(2)).`
Putting `x^(2)=v and u=sin x^(2) = sin v,` we get
`y=e^(u),u=sin v, and v=x^(2)`
`therefore" "(dy)/(du)=e^(u),(du)/(dv)=cos v, and (dv)/(dx)=2x`
`"Now, "(dy)/(dx)xx(dy)/(du)xx(du)/(dv)xx(dv)/(dx)`
`=e^(u)cos v 2=e^(sin v)cos v 2x`
`=e^(sin x^(2))cos x^(2)2x`