How do I find the derivative of $​y=\sin(\tan x^2)$

Question

How do I find the derivative of $y=\sin(\tan\,x^2)$

I used the chain rule

$\dfrac{\mathrm d}{\mathrm dx}(f(\sin(\tan\,x^2)))=f'(\sin(\tan\,x^2))×g'(\tan\,x^2)$

But don't know how to solve its further steps, should I apply chain rule again in the first part. If yes then how it is possible, if no then how to solve it.

Answer 1

With due acknowledgement to @Friedrich Philipp's comment, the functions $f$, $g$ and $h$ in the chain rule are here $f(x) = \sin(x)$, $g(x)=\tan(x)$ and $h(x)=x^2$. You have $y=f(g(h(x)))$. The chain rule, applied twice, says $y′=f′(g(h(x)))\cdot g'(h(x)) \cdot h'(x)$.We have

$$ f'(x) = (\sin(x))' = \cos(x) $$ $$ g'(x) = (\tan(x))' = \sec^2(x) $$ $$ h'(x) = (x^2)' = 2x $$

so $y' = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) = \cos(\tan(x^2)) \cdot \sec^2(x^2) \cdot 2x $

Answer 2

$$\begin{equation}\begin{split}\dfrac{\mathrm d}{\mathrm dx}(f(\sin(\tan\,x^2)))&=f'(\sin(\tan\,x^2))×g'(\tan\,x^2)\\&=\cos(\tan\,x^2)\cdot\dfrac{\mathrm d}{\mathrm dx}(tan\,x^2)\\&=\cos(\tan\,x^2)\sec^2(x^2)×2x\end{split}\end{equation}$$

Here is the similar problem with full explanation

derivatives of trig functions

Answer 3

You have to use the chain rule twice: $$\frac{\mathrm d}{\mathrm dx}f(g(h(x)) = f'(g(h(x)))\frac{\mathrm d}{\mathrm dx}g(h(x)) = f'(g(h(x)))g'(h(x))\frac{\mathrm d}{\mathrm dx}h(x) = f'(g(h(x)))g'(h(x))h'(x)$$ In your case, \begin{align} f(x) &= \sin x && \implies & f'(x) &= \cos x\\ g(x) &= \tan x && \implies & g'(x) &= 1+\tan^2 x\\ h(x) &= x^2 && \implies & h'(x) &= 2x \end{align} Therefore in total you get $$\frac{\mathrm d}{\mathrm dx}\sin(\tan x^2) = \cos(\tan x^2)\left(1+\tan^2(x^2)\right)2x$$