Find the values
Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$ in each of the following:

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

Solution:

We are given with an equation $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have to find $\frac{d y}{d x}$ of it, so by differentiating the equation on both sides with respect to $x$, we get,

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

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