How do you differentiate #f(x)=(sin2x^2)/4# using the chain rule?

1 Answer
Feb 9, 2016

Answer:

#f'(x)=xcos(2x^2)#

Explanation:

According to the chain rule, since the derivative of #sin(x)# is #cos(x)#, we will differentiate the outside #sin# function to give us #cos#, and then multiply that by the derivative of the inside function, which is #2x^2#.

Formally, this can be written as

#d/dx[sin(g(x))]=cos(g(x))*g'(x)#

Here, notice first that #sin(2x^2)# is being multiplied by #1"/"4#, which we can bring out of the expression:

#f'(x)=1/4cos(2x^2)*d/dx[2x^2]#

Since the derivative of #2x^2# is #4x#, this gives us

#f'(x)=1/4cos(2x^2)*4x#

The #4#s will cancel, leaving us with

#f'(x)=xcos(2x^2)#