How do you use the chain rule to differentiate #f(x)=sin(cos(tan(x^3+sin(x^2))))#?

Doing the chain rule is a lot like unwrapping a present: you need to peel back layer by layer till you get to the core.

How do you go about this? Well, you take a derivative of each outmost function without touching any of the inner functions, and then do the same for the next most outer function (it's hard to explain in just words).

So, let's look at #f(x)# here: the outermost function I see here is a #sin(x)#. The derivative of #sin(x) = cos(x)#, so our first step would be:

#=> cos(cos(tan(x^3+sin(x^2))#

So we've peeled back the first layer. The secondmost outer function is #cos(x)#. The derivative of #cos(x) = -sin(x)#, so:

#=> [cos(cos(tan(x^3+sin(x^2))] * [-sin(tan(x^3+sin(x^2))]#

The thirdmost outer function is tan(x). The derivative of #tan(x)# is #sec^2(x)#, so:

#=> [cos(cos(tan(x^3+sin(x^2))] * [-sin(tan(x^3+sin(x^2))] * [sec^2(x^3+sin(x^2))]#

The innermost function (our final layer) - is #x^3+sin(x^2)#

Now notice here that while #x^3# has a very straightforward derivative of #3x^2#, #sin(x^2)# will require a chain rule of it's own to solve. We know that the derivative of #sin(x) = cos(x)#, so we'd have:

#=> cos(x^2) * d/dx(x^2)#
# = 2xcos(x^2)#

Now we just put everything back together, clean up some stuff, and we have our final answer:

#f'(x) = - [3x^2 + 2xcos(x^2)] [cos(cos(tan(x^3+sin(x^2))] * [sin(tan(x^3+sin(x^2))] * [sec^2(x^3+sin(x^2))] * #

Notice how we've peeled back our original expression layer by layer. This is the crux of how to solve chain rule problems: identify your functions, and systematically take derivatives until you reach the core.

I've got a video that does a better job of explaining the concept of the chain rule. You can check it out here.

If you'd like some more practice, I have a practice problem video that you can check out here .

Hope that helped :)