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Question:

$y=\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}$, show that $\frac{d y}{d x}=\sec x(\tan x+\sec x)$

Solution:

$y=\sqrt{\frac{\sec x-\tan x}{\sec x+\tan x}}$

$y=\sqrt{\frac{\frac{1}{\cos x} \frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}}=\sqrt{\frac{1-\sin x}{1+\sin x}}$

$u=1-\sin x, v=1+\sin x, z=\frac{1-\sin x}{1+\sin x}$

Formula : $\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}$

According to quotient rule of differentiation

If $z=\frac{u}{v}$

$\mathrm{dz} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$

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

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

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

According to the chain rule of differentiation

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{dy}}{\mathrm{dz}} \times \frac{\mathrm{dz}}{\mathrm{dx}}$

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

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

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

(Multiplying and dividing by $\left.(1+\sin x)^{\frac{3}{2}}\right)$

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

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

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

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

$=\left[\cos x \times(1+\sin x)^{1}\right] \times(\cos x)^{-3}$

$=\left[(1+\sin x)^{1}\right] \times(\cos x)^{-3+1}$

$=\frac{1+\sin x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{1} x} \times \frac{1+\sin x}{\cos ^{1} x}$

$=\operatorname{secx}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)$

$=\operatorname{secx}(\sec x+\tan x)$

HENCE PROVED