How do you differentiate #f(x) = sin(xcos(x))# using the chain rule?

1 Answer
Dec 9, 2015

Answer:

You need the product rule too. The answer is #f'(x)=cos(x cos(x)) * (cos(x)-x sin(x))#

Explanation:

The chain rule says #d/dx(g(h(x)))=g'(h(x)) * h'(x)#. For the given function #f(x)=sin(x cos(x))# we have #g(x)=sin(x)# and #h(x)=x cos(x)#.

The product rule, applied to #h(x)#, says #h'(x)=d/dx(x) * cos(x)+x * d/dx(cos(x))=cos(x)-x sin(x)#. Also, #g'(x)=cos(x)#.

Therefore,

#f'(x)=g'(h(x)) * h'(x)=cos(x cos(x)) * (cos(x)-x sin(x))#