Find the derivative of the function ${\cos }^{2}x$, by using the first principle w.r to x.

Let $\text{ Let }f\left(x\right)={\cos }^{2}x,f(x+h)={\cos }^2(x+h)$

By the first principle

$f^{\mathrm{\prime }}\left(x\right)={\mathrm{Lt}}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$={\mathrm{Lt}}_{h\to 0}\frac{{\mathrm{Cos}}^2(x+h)-{\mathrm{Cos}}^{2}x}{h}$

$\underset{h\to 0}{\mathrm{Lt}}\frac{\mathrm{Sin}(x+x+h)\mathrm{Sin}(x-x-h)}{h}$

$=\underset{h\to 0}{Lt}\frac{\mathrm{Sin}(x+x+h)\mathrm{Sin}(x-x-h)}{h}$

$\boxed{\because {\mathrm{Cos}}^{2}B-{\mathrm{Cos}}^{2}A=\mathrm{Sin}(A+B)\mathrm{Sin}(A-B)}$

$=\underset{h\to 0}{Lt}\frac{\mathrm{Sin}(2x+h)\mathrm{Sin}(-h)}{h}=-\mathrm{Sin}2x\underset{h\to 0}{Lt}\frac{\mathrm{Sin}h}{h}$

$\boxed{f^1\left(x\right)=-\mathrm{Sin}2x}$