#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 3

$-\left(\frac{y}{x}\right)^{\frac{1}{3}}$

Hint:

Use the differentiation formula of $\left(x^{n}\right)$

i.e. $\frac{d\left(x^{n}\right)}{d x}=n x^{n-1}$

Given:

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

Solution:

Differentiate the given equation w.r.t x

$\frac{d}{d x}\left(x^{\frac{2}{3}}+y^{\frac{2}{3}}\right)=\frac{d}{d x}\left(a^{\frac{2}{3}}\right)$

$\frac{d}{d x}\left(x^{\frac{2}{3}}\right)+\frac{d}{d x}\left(y^{\frac{2}{3}}\right)=0 \quad\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$\frac{2}{3}(x)^{\frac{2}{3}-1}+\frac{d\left(y^{\frac{2}{3}}\right)}{d y} \times \frac{d y}{d x}=0 \quad\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\frac{2}{3}(x)^{\frac{-1}{3}}+\frac{2}{3}(y)^{\frac{2}{3}-1} \frac{d y}{d x}=0$

$\frac{2}{3}(x)^{\frac{-1}{3}}+\frac{2}{3}(y)^{\frac{-1}{3}} \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{\frac{-2}{3}(x)^{\frac{-1}{3}}}{\frac{2}{3}(y)^{\frac{-1}{3}}}=\frac{-(x)^{\frac{-1}{3}}}{(y)^{\frac{-1}{3}}}$

$\frac{d y}{d x}=-\frac{(y)^{\frac{1}{3}}}{(x)^{\frac{1}{3}}} \quad\left[\because(x)^{-n}=\frac{1}{(x)^{n}}\right]$

$\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$

Hence  $\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$is required answer