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

Question

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 1

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

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

Answer 2

Your answer is not correct because $g' \neq 3\tan^2u$. For getting $g'$ you need to apply the chain rule once more. $$g(u) = f(h(u)) = f(\tan u) = \tan^3u$$ where $f(x) = x^3$ and $h(u) = \tan u$.

So $g'(u) = f'(h(u))*h'(u) = 3(h(u))^2*\sec u = 3\tan^2u * \sec u$

Answer 3

$$u'=2x$$ $$g'=3\tan^2u \cdot sec^2u$$

$$f'(x)=2x \cdot 3\tan^2(x^2)\sec^2(x^2) = 6x\tan^2(x^2)\sec^2(x^2)$$