What is the derivative of #f(x)=cos(sin(5x))#?

1 Answer
Nov 13, 2015

#f'(x)=-5*sin(sin(5x))*cos(5x)#

Explanation:

To put it in simple words, when using the chain rule we differentiate what's outside, all that multiplied by what's inside (called function composition). Before solving your exercise, here's a little example:

#g(x)=sin(2x)#
#g'(x)=2cos(2x)#

So we can see the outside is the trigonometric function, and the inside is #2x#. The problem follows the same path, just a bit deeper, like inception. So:

#f(x)=cos(sin(5x))#
#f'(x)=d/dx(cos(sin(5x))#

The derivative of #cos(sin(5x))# is #-sin(sin(5x))#

Now, the differentiation of the inside, #sin(5x)#, is #cos(5x)# and the differentiation of what's inside is 5. Multiply all that and you get

#f'(x)=-sin(sin(5x))*cos(5x)*5#

If you want to rearrange:
#f'(x)=-5*sin(sin(5x))*cos(5x)#