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

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.

• Use that $\frac{\sin x^2}{x}=\frac{\sin x^2}{x^2}\cdot \frac{x^2}{x}-$ – mfl Dec 4 '16 at 12:13
• You have a limit of a product. Turn it into a product of the limits. – Patrick Stevens Dec 4 '16 at 12:18
• There was a typo in the $\cos$ term ($t\,x$ instead of $x$.) I have edited it. – Julián Aguirre Dec 4 '16 at 12:19
• Yes, thank you. There was a typo alright – SAK Dec 4 '16 at 12:55

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

• Thank you. But as t will tend to zero, we get 2xcos(x$^2$+x)...... – SAK Dec 4 '16 at 13:12
• I correct it @SAK – Khosrotash Dec 4 '16 at 13:47
• Thank you very much.....it now seems so simple!! – SAK Dec 4 '16 at 14:32

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.