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 '16 at 12:13
  • $\begingroup$ You have a limit of a product. Turn it into a product of the limits. $\endgroup$ – Patrick Stevens Dec 4 '16 at 12:18
  • $\begingroup$ There was a typo in the $\cos$ term ($t\,x$ instead of $x$.) I have edited it. $\endgroup$ – Julián Aguirre Dec 4 '16 at 12:19
  • $\begingroup$ Yes, thank you. There was a typo alright $\endgroup$ – SAK Dec 4 '16 at 12:55

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

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


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.