Question
Differentiate the following functions with respect to x:
$\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
$\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
✓
Answer
Let $\text{y}=\sqrt{\frac{1+\sin\text{x}}{1-\sin\text{x}}}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}\Big[\frac{(1-\sin\text{x})(\cos\text{x})-(1+\sin\text{x})(-\cos\text{x})}{(1-\sin\text{x})^2}\Big]$
$=\frac{\text{1}}{\text{2}}\frac{(1+\sin\text{x})^\frac{1}{2}}{(1-\sin\text{x})^\frac{1}{2}}\Big[\frac{\cos\text{x}-\cos\text{x}+\cos\text{x}\sin\text{x}+\sin\text{x}\cos\text{x}}{(1-\sin\text{x})^2}\Big]$
$=\frac{1}{2}\times\frac{2\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}\sqrt{1-\sin\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\sqrt{1-\sin^2\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\cos\text{x}(1-\sin\text{x})}$ $\big[\text{Using }1-\sin^2\text{x}=\cos^2\text{x}\big] $
$=\frac{1}{(1-\sin\text{x})}\times\frac{(1+\sin\text{x})}{(1+\sin\text{x})}$
Thus, $ \frac{\text{dy}}{\text{dx}}=\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos^2\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\sec^2\text{x}+\tan\text{x}\sec\text{x}$
$\frac{\text{dy}}{\text{dx}}=\sec\text{x}[\tan\text{x}+\sec\text{x}] $
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)$
$=\frac{\text{1}}{\text{2}}\Big(\frac{1+\sin\text{x}}{1-\sin\text{x}}\Big)^{\frac{1}{2}}\Big[\frac{(1-\sin\text{x})(\cos\text{x})-(1+\sin\text{x})(-\cos\text{x})}{(1-\sin\text{x})^2}\Big]$
$=\frac{\text{1}}{\text{2}}\frac{(1+\sin\text{x})^\frac{1}{2}}{(1-\sin\text{x})^\frac{1}{2}}\Big[\frac{\cos\text{x}-\cos\text{x}+\cos\text{x}\sin\text{x}+\sin\text{x}\cos\text{x}}{(1-\sin\text{x})^2}\Big]$
$=\frac{1}{2}\times\frac{2\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}(1-\sin\text{x})^\frac{3}{2}}$
$=\frac{\cos\text{x}}{\sqrt{1+\sin\text{x}}\sqrt{1-\sin\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\sqrt{1-\sin^2\text{x}}(1-\sin\text{x})}$
$=\frac{\cos\text{x}}{\cos\text{x}(1-\sin\text{x})}$ $\big[\text{Using }1-\sin^2\text{x}=\cos^2\text{x}\big] $
$=\frac{1}{(1-\sin\text{x})}\times\frac{(1+\sin\text{x})}{(1+\sin\text{x})}$
Thus, $ \frac{\text{dy}}{\text{dx}}=\frac{1}{\cos^2\text{x}}+\frac{\sin\text{x}}{\cos^2\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\sec^2\text{x}+\tan\text{x}\sec\text{x}$
$\frac{\text{dy}}{\text{dx}}=\sec\text{x}[\tan\text{x}+\sec\text{x}] $
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