Question

Find the derivative of $\sin (x^2+1)$ with respect to $x$ from first principle. (IIT-JEE, 1978)

Answer 1

Let $F\left(x\right)=\sin (x^2+1),then,f(x+h)=\sin [{(x+h)}^2+1]$

$\therefore \underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{k\to 0}{lim}\frac{\sin [{(x+h)}^2+1]-\sin [x^2+1]}{h}$

$\Rightarrow f^{{}^{\prime }}\left(x\right)=\underset{h\to 0}{lim}2\cos \left(\frac{2x^2+h^2+2xh+2}{2}\right)\times \frac{\sin \left(\frac{h^2+2xh}{2}\right)}{h}$

$=2\cos (x^2+1)\underset{h\to 0}{lim}\frac{\sin \left[\frac{h^2+2xh}{2}\right]}{h\left[\frac{h+2x}{2}\right]}\left(\frac{h+2x}{2}\right)$

$=2x\cos (x^2+1)$