How do I find the derivative of $​y=\sin(\tan x^2)$ 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.
 A: 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 $
A: $$\begin{equation}\begin{split}\dfrac{\mathrm d}{\mathrm dx}(f(\sin(\tan\,x^2)))&=f'(\sin(\tan\,x^2))×g'(\tan\,x^2)\\&=\cos(\tan\,x^2)\cdot\dfrac{\mathrm d}{\mathrm dx}(tan\,x^2)\\&=\cos(\tan\,x^2)\sec^2(x^2)×2x\end{split}\end{equation}$$
Here is the similar problem with full explanation
derivatives of trig functions
A: 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
$$\frac{\mathrm d}{\mathrm dx}\sin(\tan x^2) =
\cos(\tan x^2)\left(1+\tan^2(x^2)\right)2x$$
