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  • Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 24

Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 24

Answers (1)

Answer:

                (a)

Hint:

We must know about the derivative of logarithm.

Given:

                y=x^{n-1} \log x

Explanation:

                  y=x^{n-1} \log x

                 \\ \frac{d y}{d x}=y_{1}=(n-1) x^{n-2} \log x+\frac{x^{n-1}}{x} \\

                \begin{aligned} & &y_{1}=\frac{(n-1) x^{n-1} \log x+x^{n-1}}{x} \end{aligned}

                x y_{1}=(n-1) y+x^{n-1} \\                                                                                                     \left[\because y=x^{n-1} \log x\right]

                x y_{2}+y_{1}=(n-1) y_{1}+(n-1) x^{n-2}

                \begin{aligned} & \\ &x y_{2}+y_{1}=(n-1) y_{1}+\frac{(n-1) x^{n-1}}{x} \end{aligned}                                                                                                      

                x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x^{n-1}          

                \\ x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1)\left[x y_{1}-(n-1) y\right] \\                                                                                                                     

                x^{2} y_{2}+x y_{1}=x(n-1) y_{1}+(n-1) x y_{1}-(n-1)^{2} y \\

                \begin{aligned} & &x^{2} y_{2}+x y_{1}=2 x(n-1) y_{1}-(n-1)^{2} y \end{aligned}

                x^{2} y_{2}+x y_{1}-2 x(n-1) y_{1}=-(n-1)^{2} y \\

                x^{2} y_{2}+x y_{1}(1-2 n+2)=-(n-1)^{2} y \\

                \begin{aligned} & &x^{2} y_{2}+(3-2 n) x y_{1}=-(n-1)^{2} y \end{aligned}

 

Posted by

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