How do you differentiate #f(x)=e^(4x)*sin(5-x^2)# using the product rule?

Question

1213 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-04T03:44:52

#f'(x)=2e^(4x)(2sin(5-x^2)-xcos(5-x^2))#

#f'(x)=sin(5-x^2)d/dx[e^(4x)]+e^(4x)d/dx[sin(5-x^2)]#

Find each individual derivative, both of which use the chain rule:

#d/dx[e^(4x)]=e^(4x)d/dx[4x]=4e^(4x)#

#d/dx[sin(5-x^2)]=cos(5-x^2)d/dx[5-x^2]=-2xcos(5-x^2)#

Plug back in:

#f'(x)=4e^(4x)sin(5-x^2)-2xe^(4x)cos(5-x^2)#

#f'(x)=2e^(4x)(2sin(5-x^2)-xcos(5-x^2))#