Question

How do you differentiate `f(x)=cos(e^(3x^2)+7)` using the chain rule?

Answer 1

`f^'=-6xe^(3x^2)sin(e^(3x^2)+7)`

Explanation:

I prefer to use the form `(dy)/(dx)`

Let `u=e^(3x^2)+7=>" "(du)/(dx)=6xe^(3x^2)`

Write `f(x)" as " y=cos(u)=>" "(dy)/(du)=-sin(u)`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
But `(dy)/(dx)=(dy)/(cancel(du))xx(cancel(du))/(dx)`

`=> (dy)/(dx)=(6xe^(3x^2))(-sin(e^(3x^2)+7))`

`=> (dy)/(dx)=-6xe^(3x^2)sin(e^(3x^2)+7)`

Or if you prefer `f^'=-6xe^(3x^2)sin(e^(3x^2)+7)`

As confirmation: