Derivative of $\sin(x^2)$ using first principle

Question

I am able to find derivatives of $\sin x$ and $\sin 2x$ using first principle (Using the formula for $\sin(A)-\sin(B)$ and subsequently using $\lim_{x\rightarrow 0}$ $\frac{\sin x}{x}$ = 1. But I am getting stuck in trying to find Derivative of $\sin(x^2)$ using the same.

After using the Sin A - Sin B formula I get the following result but then I am unable to separate out $x$ and $t$ to get a $\frac{\sin(t)}{t}$ form: $$\frac{2\cos(x^2+x\,t+\frac{t^2}{2})\sin(x\,t+\frac{t^2}{2})}{t}$$ and solve it further.

Request Guide.

Answer 1

$$\lim_{t \to 0}\frac{\sin((x+t)^2)-\sin(x^2)}{t}=$$

$$\lim_{t \to 0}\frac{2\sin(\frac{(x+t)^2-x^2}{2})\cdot \cos( \frac{(x+t)^2+x^2}{2})}{t}=$$

$$\lim_{t \to 0}\frac{2\sin(\frac{2xt+t^2}{2})\cdot \cos\frac{(x+t)^2+x^2))}{2}}{t}\times \frac{\frac{2x+t}{2}}{\frac{2x+t}{2}}=\\ $$ $$\lim_{t \to 0}2\frac{\sin(\frac{2xt+t^2}{2})}{\frac{2xt+t^2}{2}}\times \frac{\frac{2x+t}{2}}{1}\times \cos\frac{(x+t)^2+x^2}{2}=\\ \lim_{t \to 0}2\times 1\times \frac{\frac{2x+t}{2}}{1}\times \cos\frac{(x+t)^2+x^2}{2}=\\2\times \frac{2x+0}{2} \times \cos\frac{(x+0)^2+x^2}{2}=\\2 \times x\times \cos(x^2)$$

Answer 2

You are right to be stuck as the transformation is not totally obvious.

Notice that

$$\frac{\sin(xt+\dfrac{t^2}2)}t=\frac{\sin(xt+\dfrac{t^2}2)}{xt+\dfrac{t^2}2}\frac{xt+\dfrac{t^2}2}{t}=\frac{\sin(xt+\dfrac{t^2}2)}{xt+\dfrac{t^2}2}\left(x+\dfrac t2\right).$$

Then as the argument of the sine tends to zero, the limit of this expression is just $1\cdot x$. Now the original limit should be doable.