Differentiate `log{(a+bsinx)/(a-bsinx)}` with respect to `x` .

Explanation

Let,

`y=log{(a+bsinx)/(a-bsinx)}`

`y=log(a+bsinx)-log(a-bsinx)`

`frac{dy}{dx}=frac{1}{a+bsinx} *frac{d}{dx} (a+bsinx)+frac{1}{a-bsinx} *frac{d}{dx} (a-bsinx)`

`frac{dy}{dx}=frac{1}{a+bsinx} (bcosx)-frac{1}{a-bsinx}* (-bcosx)`

`frac{dy}{dx}=(bcosx){frac{1}{a+bsinx} +frac{1}{a-bsinx}}`

`frac{dy}{dx}=(bcosx){frac{a-bsinx+a+bsinx}{a^2-b^2sin^2x}}`

`frac{dy}{dx}=(bcosx){frac{2a}{a^2-b^2sin^2x}}`

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