#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.4 question 5 maths textbook solution

$\frac{-b^{2} x}{a^{2} y}$

Hint:

Use chain rule and differentiation formulas like $\frac{d x^{n}}{d x}=n x^{n-1}$

Given:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Solution:

Differentiate the given equation w.r.t x

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

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

$\frac{1}{a^{2}} \cdot \frac{d\left(x^{2}\right)}{d x}+\frac{1}{b^{2}} \cdot \frac{d\left(y^{2}\right)}{d x}=0$

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

$\frac{2 x}{a^{2}}+\frac{1}{b^{2}}\left(2 y \cdot \frac{d y}{d x}\right)=0$

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

$\frac{d y}{d x}=\frac{\frac{-2 x}{a^{2}}}{\frac{2 y}{b^{2}}}=\frac{-2 x \times b^{2}}{2 y \times a^{2}}$

$\frac{d y}{d x}=\frac{-x b^{2}}{y a^{2}}$

$\frac{d y}{d x}=-\frac{b^{2} x}{a^{2} y}$

Hence $\frac{d y}{d x}=-\frac{b^{2} x}{a^{2} y}$  is the required answer