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  • Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 7

Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 7

Answers (1)

Answer:

        \frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}}

Hint:

You have to know about derivative of

        \frac{ \log x}{x}

Given:

I\! f\: \:y= \frac{ \log x}{x}, show\: \: that

\frac{d^{2} y}{d x^{2}}=\frac{2 \log x-3}{x^{3}}

Solution:

Let \:y= \frac{ \log x}{x}

Use quotient rule

        \begin{aligned} &\frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \\ &u=\log x\: \&\: v=x \\ &\frac{d y}{d x}=\frac{\log x}{x} \\ &\frac{d y}{d x}=\frac{x \frac{d}{d x} \log x-\log x \frac{d}{d x} x}{x^{2}} \quad\left(\frac{d \log x}{d x}=\frac{1}{x}\right) \\ &\frac{d y}{d x}=\frac{x \frac{1}{x}-\log x \cdot 1}{x^{2}} \\ &\frac{d y}{d x}=\frac{1-\log x}{x^{2}} \end{aligned}

Use Quotient rule again

        \begin{aligned} &\frac{u}{v}=\frac{u^{1} v-v^{1} u}{v^{2}} \quad \\ &u=1-\log x \: \&\: v=x^{2} \\ &\frac{d^{2} y}{d x^{2}}=\frac{x^{2} \frac{d}{d x}(1-\log x)-\frac{d}{d x} x^{2}(1-\log x)}{\left(x^{2}\right)^{2}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{x^{2}\left(\frac{-1}{x}\right)-(1-\log x) 2 x}{x^{4}} \quad\left(\frac{d-\log x}{d x}=\frac{-1}{x}, \frac{d}{d x} x^{2}=2 x\right) \end{aligned}

        \begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-x-2 x+2 x \log x}{x^{4}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 x \log x-3 x}{x^{4}} \\ &\frac{d^{2} y}{d x^{2}}=\frac{2 x \log x-3}{x^{3}} \end{aligned}

Hence proved

Posted by

Gurleen Kaur

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