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Question Number 141769 by 7770 last updated on 23/May/21

$$\mathbf{If}\mathbf{y}=2\mathbf{sinx}+\mathbf{tanx},\mathbf{prove}\mathbf{that}$$$$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}=2\mathbf{sinx}({\mathbf{sec}}^3\mathbf{x}-1)$$$$\mathbf{Any}\mathbf{detailed}\mathbf{solution}\mathbf{please}.$$

Answered by physicstutes last updated on 23/May/21

$$y=2\sin x+\tan x$$$$\frac{dy}{dx}=2\cos x+{\sec }^{2}x$$$$\frac{d^{2}y}{{dx}^2}=-2\sin x+\frac{\left({\cos }^{2}x\right)\left(0\right)-\left(1\right)\left(2\cos x\right)(-\sin x)}{{\cos }^{4}x}$$$$\frac{d^{2}y}{{dx}^2}=\frac{2\sin x\cos x}{{\cos }^{4}x}-2\sin x=\frac{2\sin x}{{\cos }^{3}x}-2\sin x=2\sin x(\frac{1}{{\cos }^{3}x}-1)=2\sin x({\sec }^{3}x-1)$$$$\mathrm{as}\mathrm{required}.$$

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