Find dy dx in the following cases: x^2 a^2 + y^2 b^2 =1 - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$

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Answer

Given,
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$
Differentiating with resepct to x,
$\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}\Big)=\frac{\text{d}}{\text{dx}}(1)$
$\Rightarrow\frac{\text{d}}{\text{dx}}\Big(\frac{\text{x}^2}{\text{a}^2}\Big)+\frac{\text{d}}{\text{dx}}\Big(\frac{\text{y}^2}{\text{b}^2}\Big)=0$
$\Rightarrow\frac{1}{\text{a}^2}(2{\text{x}})+\frac{1}{\text{b}^2}(2\text{y})\frac{\text{d}}{\text{dx}}=0$
$\Rightarrow\frac{2\text{y}}{\text{b}^2}\frac{\text{dy}}{\text{dx}}=-\frac{2{\text{x}}}{\text{a}^2}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\Big(\frac{2{\text{x}}}{\text{a}^2}\Big)\Big(\frac{\text{b}^2}{2\text{y}}\Big)$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{{\text{b}^2\text{x}}}{\text{a}^2\text{y}}$

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