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

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Answer 1 · 2016-07-10T19:14:25

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

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

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