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InterviewSolution
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If \(\rm f(x)=\tan^{-1} \left[\dfrac{\sin x}{1 + \cos x}\right]\), then what is the first derivative of f(x)?1. 1/22. -1/23. 24. -2 |
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Answer» Correct Answer - Option 1 : 1/2 Concept: Trigonometric Identities: sin2 θ + cos2 θ = 1. sin 2θ = 2 sin θ cos θ. cos 2θ = cos2 θ - sin2 θ.
Calculation: Let us express \(\rm\dfrac{\sin x}{1 + \cos x}\) in terms of tan x. \(\rm \dfrac{\sin x}{1 + \cos x}=\dfrac{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}{\left (\cos^2 \tfrac{x}{2}+\sin^2 \tfrac{x}{2} \right ) + \left (\cos^2 \tfrac{x}{2}-\sin^2 \tfrac{x}{2} \right )}\) = \(\rm \dfrac{2\sin \tfrac{x}{2}\cos \tfrac{x}{2}}{2\cos^2 \tfrac{x}{2}}= \dfrac{\sin \tfrac{x}{2}}{\cos \tfrac{x}{2}}=\tan \dfrac{x}{2}\). ∴ \(\rm f(x)=\tan^{-1} \left[\dfrac{\sin x}{1 + \cos x}\right]= \tan^{-1}\left (\tan \dfrac{x}{2} \right )=\dfrac{x}{2}\). And, the first derivative of f(x) = f'(x) = \(\rm \dfrac{d}{dx}\left (\dfrac{x}{2} \right )=\dfrac{1}{2}\). |
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