Question

Find the derivative of $\sin x$ with respect to $x$ from first principles.

Answer 1

First principle of differentiation :$\frac{dy}{dx}={lim}_{\delta x\to 0}\frac{f(x+\delta x)-f\left(x\right)}{\delta x}$

Here $f\left(x\right)=\sin x$

$\Rightarrow f(x+\delta x)=\sin (x+\delta x)$

$\Rightarrow f(x+\delta x)-f\left(x\right)=\sin (x+\delta x)-\sin x$

We know that $\text{sin}C-\text{sin}D=2\text{cos}\left(\frac{C+D}{2}\right)\text{sin}\left(\frac{C-D}{2}\right)$

$\Rightarrow f(x+\delta x)-f\left(x\right)=2\cos \left(\frac{x+\delta x+x}{2}\right)\sin \left(\frac{\delta x}{2}\right)$

$\Rightarrow \frac{dy}{dx}={lim}_{\delta x\to 0}\frac{2\cos (x+\frac{\delta x}{2})\sin \left(\frac{\delta x}{2}\right)}{\delta x}$

$\Rightarrow \frac{dy}{dx}={lim}_{\delta x\to 0}\cos (x+\frac{\delta x}{2})\frac{\sin \left(\frac{\delta x}{2}\right)}{\frac{\delta x}{2}}$

$\Rightarrow \frac{d\left(\sin x\right)}{dx}=\cos x$ as ${lim}_{x\to 0}\frac{\sin x}{x}=1$