#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 15

$2 \frac{d y}{d x}+y^{3}=0$

Hint:

Use product rule and $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$

Given:

$x y^{2}=1$

Solution:

Differentiate $\left[x y^{2}=1\right]$  w.r.t x

$\frac{d\left(x y^{2}\right)}{d x}=\frac{d(1)}{d x}$

$x \frac{d y^{2}}{d x}+y^{2} \frac{d x}{d x}=0$                $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$    [Product rule $\frac{d(u v)}{d x}=u \cdot \frac{d v}{d x}+v \cdot \frac{d u}{d x}$]

$x \frac{d y^{2}}{d y} \times \frac{d y}{d x}+y^{2}=0$

$x(2 y) \times \frac{d y}{d x}+y^{2}=0$                        $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$2 x y \frac{d y}{d x}+y^{2}=0$

\begin{aligned} &\frac{d y}{d x}=-\frac{y^{2}}{2 x y}=\frac{-y}{2 x} \\ &2 \frac{d y}{d x}=-\frac{y}{x} \end{aligned}

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

$2 \frac{d y}{d x}+y^{3}=0$

Hence, if  $x y^{2}=1$  then $2 \frac{d y}{d x}+y^{3}=0$  is the required answer

Hence proved