Question

Find from first principles the differential coefficient of $\sin 2x$.

Answer 1

We have,

$g\left(x\right)=\sin 2x$

Put $x=x+h$

$g(x+h)=\sin 2(x+h)$

We know that

$g^{\prime }\left(x\right)={lim}_{h\to 0}\frac{g(x+h)-g\left(x\right)}{h}$

Therefore,

$g^{\prime }\left(x\right)={lim}_{h\to 0}\frac{\sin 2(x+h)-\sin 2x}{h}$

$g^{\prime }\left(x\right)={lim}_{h\to 0}\frac{2\cos \frac{2x+2h+2x}{2}.\sin \frac{2x+2h-2x}{2}}{h}$

$g^{\prime }\left(x\right)={lim}_{h\to 0}\frac{2\cos (2x+h).\sin h}{h}$

$g^{\prime }\left(x\right)=2\cos (2x+0).\times 1$

$g^{\prime }\left(x\right)=2\cos 2x$

Hence, this is the answer.