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Calculus Find derivative with respect to x of the function f(x)=ln sqrt((1-sin x)/(1sin x)) at x!=+-pi//2+pi n,n inZ

Calculus Find derivative with respect to xx of the function f(x)=extln extsqrt(1−sin⁡x1sin⁡x)f(x)= ext{ln} \, ext{sqrt}\left(\frac{1-\sin x}{1\sin x}\right) at x≠±π2+πn,n∈Zx \neq \pm \frac{\pi}{2} + \pi n, n \in \mathbb{Z}

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Q. Calculus Find derivative with respect to xx of the function f(x)=extln extsqrt(1−sin⁡x1sin⁡x)f(x)= ext{ln} \, ext{sqrt}\left(\frac{1-\sin x}{1\sin x}\right) at x≠±π2+πn,n∈Zx \neq \pm \frac{\pi}{2} + \pi n, n \in \mathbb{Z}
  1. Simplify Function: Simplify the function inside the logarithm.\newlinef(x)=ln⁡(1−sin⁡x1+sin⁡x) f(x) = \ln \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) \newlinef(x)=12ln⁡(1−sin⁡x1+sin⁡x) f(x) = \frac{1}{2} \ln \left( \frac{1 - \sin x}{1 + \sin x} \right)
  2. Chain Rule Derivative: Use the chain rule to find the derivative.\newlinef′(x)=12⋅ddx(ln⁡(1−sin⁡x1+sin⁡x)) f'(x) = \frac{1}{2} \cdot \frac{d}{dx} \left( \ln \left( \frac{1 - \sin x}{1 + \sin x} \right) \right)
  3. Natural Logarithm Derivative: Use the derivative of the natural logarithm.\newlineddx(ln⁡(1−sin⁡x1+sin⁡x))=11−sin⁡x1+sin⁡x⋅ddx(1−sin⁡x1+sin⁡x) \frac{d}{dx} \left( \ln \left( \frac{1 - \sin x}{1 + \sin x} \right) \right) = \frac{1}{\frac{1 - \sin x}{1 + \sin x}} \cdot \frac{d}{dx} \left( \frac{1 - \sin x}{1 + \sin x} \right)
  4. Quotient Rule Derivative: Use the quotient rule to find the derivative of the fraction.\newlineddx(1−sin⁡x1+sin⁡x)=(1+sin⁡x)⋅(−cos⁡x)−(1−sin⁡x)⋅cos⁡x(1+sin⁡x)2 \frac{d}{dx} \left( \frac{1 - \sin x}{1 + \sin x} \right) = \frac{(1 + \sin x) \cdot (-\cos x) - (1 - \sin x) \cdot \cos x}{(1 + \sin x)^2} \newline=−cos⁡x−sin⁡xcos⁡x−cos⁡x+sin⁡xcos⁡x(1+sin⁡x)2 = \frac{-\cos x - \sin x \cos x - \cos x + \sin x \cos x}{(1 + \sin x)^2} \newline=−2cos⁡x(1+sin⁡x)2 = \frac{-2 \cos x}{(1 + \sin x)^2}
  5. Combine Results: Combine the results.\newlinef′(x)=12⋅1+sin⁡x1−sin⁡x⋅−2cos⁡x(1+sin⁡x)2 f'(x) = \frac{1}{2} \cdot \frac{1 + \sin x}{1 - \sin x} \cdot \frac{-2 \cos x}{(1 + \sin x)^2} \newline=12⋅−2cos⁡x(1−sin⁡x)(1+sin⁡x) = \frac{1}{2} \cdot \frac{-2 \cos x}{(1 - \sin x)(1 + \sin x)} \newline=−cos⁡x(1−sin⁡x)(1+sin⁡x) = \frac{-\cos x}{(1 - \sin x)(1 + \sin x)} \newline=−cos⁡x1−sin⁡2x = \frac{-\cos x}{1 - \sin^2 x} \newline=−cos⁡xcos⁡2x = \frac{-\cos x}{\cos^2 x} \newline=−1cos⁡x = -\frac{1}{\cos x} \newline=−sec⁡x = -\sec x

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