Question

Find the derivative of ${\csc }^{2}x$, by using first principle of derivatives.

Answer 1

$y={\csc }^{2}x$ .......$\left(1\right)$

Increase from $y$ to $y+\delta y$ correspondingly $x$ to $x+\delta x$ in the above equation$\left(1\right)$

$\Rightarrow y+\delta y={\csc }^2(x+\delta x)$ .....$\left(2\right)$

Eqn$\left(2\right)$-Eqn$\left(1\right)$

$\Rightarrow y+\delta y-y={\csc }^2(x+\delta x)-{\csc }^{2}x$

$\Rightarrow \delta y=\frac{1}{{\sin }^2(x+\delta x)}-\frac{1}{{\sin }^{2}x}$

$\Rightarrow \delta y=\frac{{\sin }^{2}x-{\sin }^2(x+\delta x)}{{\sin }^{2}x{\sin }^2(x+\delta x)}$

$\Rightarrow \delta =\frac{\sin (x+x+\delta x)\sin (x-x-\delta x)}{{\sin }^{2}x{\sin }^2(x+\delta x)}$ using ${\sin }^{2}A-{\sin }^{2}B=\sin (A+B)\sin (A-B)$

$\Rightarrow \delta y=\frac{\sin (x+x+\delta x)\sin (x-x-\delta x)}{{\sin }^{2}x{\sin }^2(x+\delta x)}$

$\Rightarrow \delta y=\frac{\sin (2x+\delta x)\sin (-\delta x)}{{\sin }^{2}x{\sin }^2(x+\delta x)}$

$\Rightarrow \delta y=\frac{-\sin (2x+\delta x)\sin \left(\delta x\right)}{{\sin }^{2}x{\sin }^2(x+\delta x)}$ since $\sin (-\theta )=-\sin \theta$

$\Rightarrow {lim}_{\delta x\to 0}\frac{\delta y}{\delta x}={lim}_{\delta x\to 0}\frac{-sin(2x+\delta x)\frac{\sin \delta x}{\delta x}}{{\sin }^{2}x{\sin }^2(x+\delta x)}$ by applying limits to both sides

$\Rightarrow \frac{dy}{dx}=\frac{-sin(2x+0)\times 1}{{\sin }^{2}x{\sin }^2(x+0)}$

$\Rightarrow \frac{dy}{dx}=\frac{-sin2x}{{\sin }^{2}x{\sin }^{2}x}$ since

${lim}_{\theta \to 0}\frac{\sin \theta }{\theta }=1$

$\Rightarrow \frac{dy}{dx}=\frac{-2\sin x\cos x}{{\sin }^{2}x{\sin }^{2}x}$ since $sin2x=2\sin x\cos x$

$\Rightarrow \frac{dy}{dx}=\frac{-2\cos x}{{\sin }^{3}x}$

$\therefore \frac{dy}{dx}=-2\cot x{\csc }^{2}x$