What's the problem with differentiating $y = \sin(x^2)$ by applying the limit definition of a derivative directly? I was taking the derivative of $y = \sin(x^2)$. I know that we can solve it by applying chain rule, but i tried without any rules, just like a normal method. This is what i did:
$$\frac{\sin((x + h)^2) - \sin((x)^2)}{h}$$
Is this method correct? If not, then why? Because wherever I search for the derivative of $y = \sin(x^2)$, none did like this. And also I am not able to come to the proper answer which is $2x\cos(x^2)$ through that method.
Can somebody help me!
 A: You didn't try hard enough. Just for curiosity's sake, here you go. Notice this follows the pattern of the usual proof of the derivative of $\sin$, using the addition formula for $\sin$.
First,
\begin{align*}
\lim_{h\to 0}&\frac{\sin((x+h)^2) - \sin(x^2)}h \\ &= \sin(x^2)\lim_{h\to 0}\frac{\cos(2xh+h^2)-1}h + \cos(x^2)\lim_{h\to 0}\frac{\sin(2hx+h^2)}h.
\end{align*}
Now, recall from your text that
$$\lim_{h\to 0}\frac{\sin h}h = 1,$$
so we have
$$\lim_{h\to 0}\frac{\sin(2xh+h^2)}h = \lim_{h\to 0}\frac{\sin(2xh+h^2)}{2xh+h^2}\lim_{h\to 0}\frac{2xh+h^2}h = 1\cdot 2x.$$
Similarly, you should recall from your text that
$$\lim_{h\to 0}\frac{\cos(h)-1}h = 0,$$
and so the same approach will show that the first term goes to $0$. This gives
$$\lim_{h\to 0}\frac{\sin((x+h)^2) - \sin(x^2)}h = 2x\cos(x^2),$$
as you desired.
COMMENT: This is essentially how the chain rule is proved, by the way, if you don't worry about every last technicality.
A: This is OK,
but the chain rule is easier.
Continue like this,
using $\sin(x) = x+O(x^3),
\cos(x) = 1-x^2/2+O(x^2)$
for small $x$:
$\begin{array}\\
\Delta_h(\sin(x^2))
&=\dfrac{\sin((x + h)^2) - \sin((x)^2)}{h}\\
&=\dfrac{\sin(x^2+2hx+h^2) - \sin(x^2)}{h}\\
&=\dfrac{\sin(x^2)\cos(2hx+h^2)+\cos(x^2)\sin(2hx+h^2) - \sin(x^2)}{h}\\
&=\dfrac{\sin(x^2)(\cos(2hx+h^2)-1)+\cos(x^2)\sin(2hx+h^2)}{h}\\
&=\dfrac{\sin(x^2)(\cos(2hx+h^2)-1)}{h}+\dfrac{\cos(x^2)\sin(2hx+h^2)}{h}\\
&\approx\dfrac{\sin(x^2)((1-(2hx+h^2)^2/2-1)}{h}+\dfrac{\cos(x^2)(2hx+O(h^2)}{h}\\
&=\dfrac{-\sin(x^2)((2hx+h^2)^2/2}{h}+\dfrac{\cos(x^2)(2hx+O(h^2)}{h}\\
&=-\sin(x^2)(h(2x+h)^2/2+2x\cos(x^2)+O(h)\\
&\to 2x\cos(x^2)\\
\end{array}
$
A: $\sin (x+h)^2 -\sin (x^2)=2\cos \big(\frac{ (x+h)^2+x^2}2 \big) \sin \big(\frac{ (x+h)^2-x^2}2 \big) $
$=2\cos \big(\frac{ (x+h)^2+x^2}2 \big) \sin \big(\frac{ h(h+2x)}2 \big) $
So
$\lim_{h\to 0}\frac{\sin (x+h)^2 -\sin (x^2)} h$
$=\lim_{h\to 0}\frac 1h 2\cos \big(\frac{ (x+h)^2+x^2}2 \big) \sin \big(\frac{ h(h+2x)}2 \big) $
$=2\cos(x^2)\lim_{h\to 0}\frac{
sin \big(\frac{ h(h+2x)}2 \big)}{\frac{h(h+2x)}2}\frac{ (h+2x)}2$
$=2x\cos(x^2)$
