Question

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

Answer 1

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

Explanation:

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

• `u = tanx`

• `d/(du)[u^2] = 2u` (from power rule):

`= 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

• `u = x^2`

• `d/(du)[tanu] = sec^2u`:

`= 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`