Find dy dx in the following cases: x^ 2 3 +y^ 2 3 =a^ 2 3 - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}}$ in the following cases:
$\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$

✓

Answer

We have, $\text{x}^{\frac{2}{3}}+\text{y}^{\frac{2}{3}}=\text{a}^{\frac{2}{3}}$
Differentiating it with respect to x, we get,
$\frac{\text{d}}{\text{dx}}\Big(\text{x}^{\frac{2}{3}}\Big)+\frac{\text{d}}{\text{dx}}\Big(\text{y}^{\frac{2}{3}}\Big)=\frac{\text{d}}{\text{dx}}\Big(\text{a}^{\frac{2}{3}}\Big)$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{2}{3}-1}+\frac{2}{3}\big(\text{y}\big)^{\frac{2}{3}-1}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}+\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow\frac{2}{3}\big(\text{y}\big)^{\frac{-1}{3}}\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{2}{3}\big(\text{x}\big)^{\frac{-1}{3}}\times\frac{3}{2\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{-1}{3}}}{\text{y}^{\frac{-1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}^{\frac{1}{3}}}{\text{y}^{\frac{1}{3}}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\big(\frac{\text{x}}{\text{y}}\big)^{\frac{1}{3}}$

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