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

Question:

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

$\sqrt{\sec x}$

 

Solution:

Let

$f(x)=\sqrt{\sec x}$

We need to find the derivative of f(x) i.e. f’(x)

We know that,

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

$f(x)=\sqrt{\sec x}$

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

Putting values in (i), we get

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

Now rationalizing the numerator by multiplying and divide by the conjugate

of $\sqrt{\sec (\mathrm{x}+\mathrm{h})}-\sqrt{\sec \mathrm{x}}$

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

Using the formula:

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

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

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

$=\lim _{h \rightarrow 0} \frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$

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

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

Using the formula:

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

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

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

$=2 \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$

$\times \frac{1}{2} \lim _{h \rightarrow 0} \sin \left(\frac{2 x+h}{2}\right)$

$\times \lim _{h \rightarrow 0} \frac{1}{(\cos (x+h) \cos x)(\sqrt{\sec (x+h)}+\sqrt{\sec x})}$

[Here, we multiply and divide by $\frac{1}{2}$ ]

$=2 \times \frac{1}{2} \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$

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

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

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

Putting h = 0, we get

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

$=\sin x \times \frac{1}{\cos x \cos x(\sqrt{\sec x}+\sqrt{\sec x})}$

$=\frac{\sin x}{\cos ^{2} x(2 \sqrt{\sec x})}$

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

$=\tan x \times \sec x \times \frac{1}{2 \sqrt{\sec x}}\left[\because \frac{\sin x}{\cos x}=\tan x\right] \&\left[\frac{1}{\cos x}=\sec x\right]$

$=\frac{1}{2} \tan x \sqrt{\sec x}$

Hence

$f^{\prime}(x)=\frac{1}{2} \tan x \sqrt{\sec x}$

 

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