Question 28.7.6: Differentiate √1-tan x/1+tan x....
Differentiate \sqrt{\frac{1-\tan x}{1+\tan x}} .
Step-by-Step
Let y=\sqrt{\frac{1-\tan x}{1+\tan x}} .
Putting \frac{1-\tan x}{1+\tan x}=t , we get y=\sqrt{t} and t=\frac{1-\tan x}{1+\tan x} .
\begin{aligned}\therefore \quad \frac{d y}{d t} & =\frac{1}{2} t^{-1 / 2}=\frac{1}{2 \sqrt{t}} .\\ \\\text { And, } \frac{d t}{d x} & =\frac{(1+\tan x) \cdot \frac{d}{d x}(1-\tan x)-(1-\tan x) \cdot \frac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}} \\ \\& =\frac{(1+\tan x)\left(-\sec ^{2} x\right)-(1-\tan x)\left(\sec ^{2} x\right)}{(1+\tan x)^{2}}=\frac{-2 \sec ^{2} x}{(1+\tan x)^{2}} . \\ \\\therefore \quad \frac{d y}{d x} & =\left(\frac{d y}{d t} \times \frac{d t}{d x}\right)=\frac{1}{2 \sqrt{t}} \times \frac{-2 \sec ^{2} x}{(1+\tan x)^{2}} \\ \\& =\frac{-\sec ^{2} x}{(1+\tan x)^{2}} \times \frac{\sqrt{1+\tan x}}{\sqrt{1-\tan x}}\\ \\& =\frac{-\sec ^{2} x}{(1+\tan x)^{3 / 2}(1-\tan x)^{1 / 2}} .\end{aligned}
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