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Provide solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 38

Answers (1)

Answer:

Proved

Hint:

You must know the derivative of logarithm and cos function

Given:

y=\log (1+\cos x) , \text { prove } \frac{d^{3} y}{d x^{3}}+\frac{d^{2} y}{d x^{2}} \times \frac{d y}{d x}=0

Solution:

y=\log (1+\cos x)

        \begin{aligned} &\frac{d y}{d x}=\frac{-\sin x}{1+\cos x}\\ &\frac{d^{2} y}{d x^{2}}=\frac{-\cos x-\cos ^{2} x-\sin ^{2} x}{(1+\cos x)^{2}}\\ &=\frac{-(\cos x+1)}{(1+\cos x)}\\ &=\frac{-1}{1+\cos x}\\ &\text { Again differentiating } \end{aligned}

        \begin{aligned} &\frac{d^{3} y}{d x^{3}}=\frac{-\sin x}{(1+\cos x)^{2}} \\ &\frac{d^{3} y}{d x^{3}}+\frac{\sin x}{(1+\cos x)^{2}}=0 \\ &\frac{d^{3} y}{d x^{3}}+\left(\frac{-1}{1+\cos x}\right)\left(\frac{-\sin x}{1+\cos x}\right)=0 \\ &\frac{d^{3} y}{d x^{3}}+\frac{d^{2} y}{d x^{2}} \times \frac{d y}{d x}=0 \end{aligned}

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