# 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, 2016 at 12:13
• You have a limit of a product. Turn it into a product of the limits. Dec 4, 2016 at 12:18
• This question was asked in my test and I cheated it from math.SE ಠ_ಠ Dec 7, 2021 at 12:39

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

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

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

• Thank you very much.....it now seems so simple!!
– SAK
Dec 4, 2016 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.