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.

  • $\begingroup$ Use that $\frac{\sin x^2}{x}=\frac{\sin x^2}{x^2}\cdot \frac{x^2}{x}-$ $\endgroup$
    – mfl
    Dec 4, 2016 at 12:13
  • $\begingroup$ You have a limit of a product. Turn it into a product of the limits. $\endgroup$ Dec 4, 2016 at 12:18
  • $\begingroup$ This question was asked in my test and I cheated it from math.SE ಠ_ಠ $\endgroup$ Dec 7, 2021 at 12:39

2 Answers 2


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

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

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

  • $\begingroup$ Thank you very much.....it now seems so simple!! $\endgroup$
    – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.