Question

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

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

Answer 1

$\mathrm{We}\mathrm{have},\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Differentiating with respect to x, we get,
$$\begin{gathered} \frac{d}{dx}\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}\right)=\frac{d}{dx}\left(1\right) \\ \Rightarrow \frac{d}{dx}\left(\frac{x^2}{a^2}\right)+\frac{d}{dx}\left(\frac{y^2}{b^2}\right)=0 \\ \Rightarrow \frac{1}{a^2}\left(2x\right)+\frac{1}{b^2}\left(2y\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0 \\ \Rightarrow \frac{2y}{b^2}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\frac{2x}{a^2} \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\left(\frac{2x}{a^2}\right)\left(\frac{b^2}{2y}\right) \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\frac{b^{2}x}{a^{2}y} \end{gathered}$$