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

(b)

Hint:

We must have known about the derivative of  $x$

Given:

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

Explanation:

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

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)

Now

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}