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Explain solution RD Sharma class 12 Chapter 11 Higher Order Derivatives Exercise Multiple Choice Question question 22

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We must have known about the derivative of  x


                y^{\frac{1}{n}}+y^{\frac{-1}{n}}=2 x  


                y^{\frac{1}{n}}+y^{\frac{-1}{n}}=2 x

Differentiate with respect to  x

{\left[\frac{1}{n} y^{\frac{1}{n}-1}-\frac{1}{n} y^{\frac{-1}{n}-1}\right] \frac{d y}{d x}=2} \\

\frac{1}{n} y^{-1}\left[y^{\frac{1}{n}}-y^{\frac{-1}{n}}\right] \frac{d y}{d x}=2 \\

\frac{1}{n y}\left[y^{\frac{1}{n}}-y^{\frac{-1}{n}}\right] \frac{d y}{d x}=2

\begin{aligned} \\ &{\left[y^{\frac{1}{n}}-y^{\frac{-1}{n}}\right] \frac{d y}{d x}=2 x y} \end{aligned}

Squaring both sides,

                \left[y^{\frac{2}{n}}-y^{\frac{-2}{n}}-2\right]\left(\frac{d y}{d x}\right)^{2}=4 x^{2} y^{2}                                                                                                  … (i)


Given    y^{\frac{1}{n}}+y^{\frac{-1}{n}}=2 x

Squaring both sides,

                y^{\frac{2}{n}}+y^{\frac{-2}{n}}+2=4 x^{2}

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

Putting this value in (i)

                4 x^{2} y^{2}=\left[4 x^{2}-2-2\right]\left(\frac{d y}{d x}\right)^{2} \\

                4 x^{2} y^{2}=\left[4 x^{2}-4\right]\left(\frac{d y}{d x}\right)^{2}

                \begin{aligned} \\ &x^{2} y^{2}=\left[x^{2}-1\right]\left(\frac{d y}{d x}\right)^{2} \end{aligned}

Differentiate with respect to  x ,

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

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

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

\begin{aligned} & \\ &{\left[x^{2}-1\right] y_{2}+x y_{1}=n^{2} y} \end{aligned} 

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