Question

Find the derivative of the following functions from the first principals w.r.t to $x$.
${\cos }^{2}x$

Answer 1

Applying the first principle to find the derivative of ${\cos }^{2}x$.

$\underset{h\to 0}{lim}\frac{{\cos }^2\left(x+h\right)-{\cos }^{2}x}{h}$

$=\underset{h\to 0}{lim}\frac{\left(\cos \left(x+h\right)-\cos x\right)\left(\cos \left(x+h\right)+\cos x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{\left(2\sin \left(\frac{x+h+x}{2}\right)\sin \left(\frac{x+h-x}{2}\right)\right)\left(2\cos \left(\frac{x+h+x}{2}\right)\cos \left(\frac{x+h-x}{2}\right)\right)}{h}$

$=\underset{h\to 0}{lim}\frac{4\sin \left(\frac{2x+h}{2}\right)\sin \left(\frac{h}{2}\right)\cos \left(\frac{2x+h}{2}\right)\cos \left(\frac{h}{2}\right)}{h}$

$=\underset{h\to 0}{lim}4\sin \left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\cos \left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\cos \left(\frac{h}{2}\right)\times \underset{h\to 0}{lim}\frac{\sin \left(\frac{h}{2}\right)}{h}$

$=\underset{h\to 0}{lim}4\sin \left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\cos \left(\frac{2x+h}{2}\right)\times \underset{h\to 0}{lim}\cos \left(\frac{h}{2}\right)\times \underset{h\to 0}{lim}\frac{\sin \left(\frac{h}{2}\right)}{\frac{2h}{2}}$

$=\underset{h\to 0}{lim}4\sin \left(\frac{2x+0}{2}\right)\times \underset{h\to 0}{lim}\cos \left(\frac{2x+0}{2}\right)\times \underset{h\to 0}{lim}\cos \left(0\right)\times \underset{h\to 0}{lim}\frac{1}{2}\times \frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}$

$=4\sin x\times \cos x\times 1\times \frac{1}{2}$

$=2\sin x\cos x$

$=\sin 2x$