How do you use the chain rule to differentiate #y=tan(x^2)+tan^2x#?

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Answer 1 · 2017-07-18T03:28:26

#y'(x) = color(blue)(2xsec^2(x^2) + 2sec^2xtanx#

We're asked to find the derivative

#(dy)/(dx) [y = tan(x^2) + tan^2x]#

using the chain rule. We'll first separate them into separate terms:

#y'(x) = d/(dx)[tan(x^2)] + d/(dx)[tan^2x]#

The chain rule can be used on both terms; we'll first differentiate the #tan^2x# term:

#d/(dx) [tan^2x] = d/(du)[u^2] (du)/(dx)#

where

#= d/(dx)[tan(x^2)] + 2d/(dx)[tanx]tanx#

The derivative of #tanx# is #sec^2x#:

#= d/(dx)[tan(x^2)] + 2sec^2xtanx#

We'll now use the chain rule on the second term:

#d/(dx) [tan(x^2)] = d/(du)[tanu] (du)/(dx)#

where

#= d/(dx)[x^2] sec^2(x^2) + 2sec^2xtanx#

The derivative of #x^2# is #2x# (power rule):

#= color(blue)(2xsec^2(x^2) + 2sec^2xtanx#