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

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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Gurleen Kaur

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