Find the derivation of each of the following from the first principle:

Solution:

Let

$\mathrm{f}(\mathrm{x})=\sqrt{\cos 3 \mathrm{x}}$

We need to find the derivative of $f(x)$ i.e. $f^{\prime}(x)$

We know that,

$\mathrm{f}^{\prime}(\mathrm{x})=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(\mathrm{x}+\mathrm{h})-\mathrm{f}(\mathrm{x})}{\mathrm{h}}$ …(i)

$f(x)=\sqrt{\cos 3 x}$

$\mathrm{f}(\mathrm{x}+\mathrm{h})=\sqrt{\cos 3(\mathrm{x}+\mathrm{h})}$

$=\sqrt{\cos (3 x+3 h)}$

Putting values in (i), we get

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sqrt{\cos (3 x+3 h)}-\sqrt{\cos 3 x}}{h}$

Now rationalizing the numerator by multiplying and divide by the conjugate

of $\sqrt{\cos (3 x+3 h)}-\sqrt{\cos 3 x}$

$=\lim _{h \rightarrow 0} \frac{\sqrt{\cos (3 x+3 h)}-\sqrt{\cos 3 x}}{h} \times \frac{\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x}}{\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x}}$

Using the formula:

$(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

$=\lim _{h \rightarrow 0} \frac{(\sqrt{\cos (3 x+3 h)})^{2}-(\sqrt{\cos 3 x})^{2}}{h(\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x})}$

$=\lim _{h \rightarrow 0} \frac{\cos (3 x+3 h)-\cos 3 x}{h(\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x})}$

Using the formula:

$\cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right)$

$=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{3 x+3 h+3 x}{2} \sin \frac{3 x+3 h-3 x}{2}}{h(\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x})}$

$=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{6 x+3 h}{2} \sin \frac{3 h}{2}}{h \sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x}}$

$=-2 \lim _{h \rightarrow 0} \frac{\sin \frac{3 h}{2}}{\frac{3 h}{2}} \times \frac{3}{2} \lim _{h \rightarrow 0} \sin \left(\frac{6 x+3 h}{2}\right) \times \lim _{h \rightarrow 0} \frac{1}{\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x}}$

[Here, we multiply and divide by $\frac{3}{2}$ ]

$=-2 \times \frac{3}{2} \lim _{h \rightarrow 0} \frac{\sin \frac{3 h}{2}}{\frac{3 h}{2}} \times \lim _{h \rightarrow 0} \sin \left(\frac{6 x+3 h}{2}\right) \times \lim _{h \rightarrow 0} \frac{1}{\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x}}$

$=-3 \times(1) \times \lim _{h \rightarrow 0} \sin \left(\frac{6 x+3 h}{2}\right) \times \lim _{h \rightarrow 0} \frac{1}{\sqrt{\cos (3 x+3 h)}+\sqrt{\cos 3 x}}$

$\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

Putting h = 0, we get

$=-3 \times \sin \left[\frac{6 x+3(0)}{2}\right] \times \frac{1}{\sqrt{\cos (3 x+3(0))}+\sqrt{\cos 3 x}}$

$=-3 \sin 3 x \times \frac{1}{2 \sqrt{\cos 3 x}}$

$=-\frac{3 \sin 3 x}{2(\cos 3 x)^{\frac{1}{2}}}$

Hence,

$f^{\prime}(x)=-\frac{3 \sin 3 x}{2(\cos 3 x)^{\frac{1}{2}}}$