calculus - Differentiate $tan^3(x^2)$

Differentiate $\tan^3(x^2)$

I first applied the chain rule and made $u=x^2$ and $g=\tan^3u$. I then calculated the derivative of $u$, which is $$u'=2x$$ and the derivative of $g$, which is

$$g'=3\tan^2u$$

I then applied the chain rule and multiplied them together, which gave me

$$f'(x)=2x3\tan^2(x^2)$$

Is this correct? If not, any hints as to how to get the correct answer?

Answer

You are almost there! In this case you need to apply the Chain Rule three times.

We have that $$(\tan^3(x^2))'=3\tan^2(x^2)\cdot(\tan(x^2))'=3\tan^2(x^2)\cdot\sec^2(x^2)\cdot(x^2)'=6x\tan^2(x^2)\sec^2(x^2)$$