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

1 Answer
Jul 10, 2016

Answer:

#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:
Tony B - Maple