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Let y=log⁡e(a+bsin⁡xa−bsin⁡x)y = \log_e \left(\frac{a + b \sin x}{a - b \sin x}\right), then value of dydx\frac{dy}{dx} is :

Solution

✅ Correct Option: 4

y=log⁡(a+bsin⁡x)−log⁡(a−bsin⁡x)y=\log(a+b\sin x)-\log(a-b\sin x).

dydx=bcos⁡xa+bsin⁡x+bcos⁡xa−bsin⁡x\frac{dy}{dx}=\frac{b\cos x}{a+b\sin x}+\frac{b\cos x}{a-b\sin x}

=bcos⁡x⋅2aa2−b2sin⁡2x=2abcos⁡xa2−b2sin⁡2x=b\cos x\cdot\frac{2a}{a^2-b^2\sin^2 x}=\frac{2ab\cos x}{a^2-b^2\sin^2 x}.

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