InterviewSolution
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InterviewSolution
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log(1+ sin2x) |
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Answer» \(\frac{d}{dx}\)log(1+sin2x) \(=\frac{1}{1+sin^2x}\frac{d}{dx}\)(1+sin2x) (By chain rule and \(\frac{d}{dx}\)log x \(=\frac{1}{x}\)) \(=\frac{1}{1+sin^2x}\)(2 sin x cos x) (∵ \(\frac{d}{dx}\) 1 = 0 and \(\frac{d}{dx}\) sin2 x = 2 sin x cos x) \(=\frac{sin^2x}{1+sin^2x}\) (∵ 2 sin x cos x = sin2 x) Hence, \(\frac{d}{dx}\)log(1+sin2 x) \(=\frac{sin^2x}{1+sin^2x}\) |
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