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  • Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 4 maths

Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 4 maths

Answers (1)

Answer:

        \frac{6}{x}

HInt:

You must know about derivative of x3 & log x

Given:

If y=x3log x, Prove that

        \frac{d^{4}y}{dx^{4}}=\frac{6}{x}

Solution:

Let y=x3log x

Use multiplicative rule

        As \: \: UV=UV^{1}+U^{1}V

        \begin{aligned} &\text { Where } U=x^{3} \& V=\log x \\ &\frac{d y}{d x}=x^{3} \frac{d}{d x} \log x+\frac{d}{d x} x^{3} \log x \\ &\frac{d y}{d x}=x^{3} \frac{1}{x}+3 x^{2} \log x \quad\left(\frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} x^{3}=3 x^{2}\right) \\ &\frac{d y}{d x}=x^{2}+3 x^{2} \log x \end{aligned}

Use multiplicative rule

        As \: \: UV=UV^{1}+U^{1}V

        \begin{aligned} &\text { Where } U=x^{2} \& V=\log x \end{aligned}

        \begin{aligned} &\frac{d^{2} y}{d x^{2}}=3\left(x^{2} \frac{d}{d x} \log x+\frac{d}{d x} x^{2} \log x\right)+\frac{d}{d x} x^{2} \\ &\frac{d^{2} y}{d x^{2}}=3\left(x^{2} \frac{1}{x}+2 x \log x\right)+2 x \quad\left(\frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} x^{2}=2 x\right) \\ &\frac{d^{2} y}{d x^{2}}=3 x+6 x \log x+2 x \\ &\frac{d^{2} y}{d x^{2}}=5 x+6 x \log x \end{aligned}

Use multiplicative rule

        As \: \: UV=UV^{1}+U^{1}V

        \begin{aligned} &\text { Where } U=x\: \&\: V=\log x \end{aligned}

        \begin{aligned} &\frac{d^{3} y}{d x^{3}}=\frac{d}{d x} 5 x+6\left(x \frac{d}{d x} \log x+\frac{d}{d x} x \log x\right) \\ &\frac{d^{3} y}{d x^{3}}=5+6\left(x \frac{1}{x}+\log x\right) \quad\left(\frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} 5 x=5\right) \\ &\frac{d^{3} y}{d x^{3}}=5+6+6 \log x \end{aligned}

        \begin{aligned} &\frac{d^{3} y}{d x^{3}}=11+6 \log x \\ &\left.\frac{d^{4} y}{d x^{4}}=0+\frac{6}{x} \quad \text { ( } \frac{d \log x}{d x}=\frac{1}{x}, \frac{d}{d x} 11=0\right) \\ &\frac{d^{4} y}{d x^{4}}=\frac{6}{x} \end{aligned}

Hence proved

Posted by

Gurleen Kaur

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