Question

Find the derivative of $\cos \sqrt{x}$ with respect to $x$ using the first principle.

Answer 1

Compute the derivative:

As we know, the formula for the first principle,

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left[\frac{f(x+h)-f\left(x\right)}{h}\right]$

We can write that

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

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }-\frac{2}{h}\sin \frac{\sqrt{x+h}+\sqrt{x}}{2}\sin \frac{\sqrt{x+h}-\sqrt{x}}{2}\left[\because \cos A-\cos B=-2\sin \left(\frac{A+B}{2}\right)\sin \left(\frac{A-B}{2}\right)\right]$

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\left[\sin \left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\frac{\sin \left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)\left(-\frac{2}{h}\right)\right]$

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim \sin }\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\underset{h\to 0}{\lim }\frac{\sin \left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}\underset{h\to 0}{\lim }\left(\frac{{\left(x+h\right)}^{{\displaystyle \frac{1}{2}}}-x^{{\displaystyle \frac{1}{2}}}}{h}\right)\left(-1\right)$

$f\text{'}\left(x\right)=-\sin \sqrt{x}\left(1\right)\times \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{1}{2\sqrt{x+h}}}-0}{1}\left[UseLHospitalruledifferentiatew.r.th\right]$

$f\text{'}\left(x\right)=-\frac{\sin \sqrt{x}}{2\sqrt{x}}$

Hence the required derivative is $-\frac{\sin \sqrt{x}}{2\sqrt{x}}$