If #f(x) =xe^x# and #g(x) = sin3x #, what is #f'(g(x)) #?

Answer 1

I think you meant to write this as: what is #(f(g(x)))'#? (or #d/dx(f(g(x)))#?...since it's supposed to be a Chain Rule problem). In that case, the answer is

#3e^{sin(3x)}cos(3x)+3e^{sin(3x)}cos(3x)sin(3x)#.

Let #h(x)=f(g(x))=e^{sin(3x)}sin(3x)#. We want to find #h'(x)=(f(g(x)))'=d/dx(f(g(x)))#.

The Chain Rule says that #h'(x)=f'(g(x)) * g'(x)#.

In this case, the Product Rule gives #f'(x)=e^[x}+xe^[x}# and we also know (by the Chain Rule again), that #g'(x)=cos(3x) * d/dx(3x)=3cos(3x)#.

Putting these things together leads to the final answer:

#h'(x)=3e^{sin(3x)}cos(3x)+3e^{sin(3x)}cos(3x)sin(3x)#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

To find ( f'(g(x)) ), first, find ( g'(x) ). Then substitute ( g(x) ) into ( f'(x) ).

( g'(x) = 3 \cos(3x) )

Now, substitute ( g(x) ) into ( f'(x) ):

( f'(g(x)) = e^{g(x)}(1+g(x)) )

Substitute ( g(x) = \sin(3x) ) into the above expression:

( f'(g(x)) = e^{\sin(3x)}(1+\sin(3x)) )

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.