Question

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

Answer 1

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

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

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

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

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

Divide both sides by $\delta x$ we get

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

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

$\Rightarrow {lim}_{x\to 0}\frac{\delta y}{\delta x}={lim}_{x\to 0}-\frac{\sin (2x+\delta x)\sin \delta x}{\delta x}$

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

$\Rightarrow \frac{dy}{dx}=-\frac{\sin (2x+0)\times 1\$since\${lim}_{x\to 0}\sin \delta x}{\delta x}=1$

$\therefore \frac{dy}{dx}=-\sin 2x=-2\cos x\sin x$