If x^{frac{2}{5}}+y^{frac{2}{5}}=a^{frac{2}{5 | vedclass

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(D) Given the equation: $x^{\frac{2}{5}} + y^{\frac{2}{5}} = a^{\frac{2}{5}}$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^{\frac{2}{

Vedclass · Accepted Answer

$-\sqrt[5]{\left(\frac{y}{x}\right)^3}$

Vedclass · Answer

(D) Given the equation: $x^{\frac{2}{5}} + y^{\frac{2}{5}} = a^{\frac{2}{5}}$Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^{\frac{2}{5}}) + \frac{d}{dx}(y^{\frac{2}{5}}) = \frac{d}{dx}(a^{\frac{2}{5}})$Using the power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ and the chain rule:$\frac{2}{5}x^{\frac{2}{5}-1} + \frac{2}{5}y^{\frac{2}{5}-1} \cdot \frac{dy}{dx} = 0$$\frac{2}{5}x^{-\frac{3}{5}} + \frac{2}{5}y^{-\frac{3}{5}} \cdot \frac{dy}{dx} = 0$Dividing by $\frac{2}{5}$:$x^{-\frac{3}{5}} + y^{-\frac{3}{5}} \cdot \frac{dy}{dx} = 0$$y^{-\frac{3}{5}} \cdot \frac{dy}{dx} = -x^{-\frac{3}{5}}$$\frac{dy}{dx} = -\frac{x^{-\frac{3}{5}}}{y^{-\frac{3}{5}}}$$\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{3}{5}}$$\frac{dy}{dx} = -\sqrt[5]{\left(\frac{y}{x}\right)^3}$Thus,the correct option is $D$.

If $x^{\frac{2}{5}}+y^{\frac{2}{5}}=a^{\frac{2}{5}}$,then $\frac{dy}{dx} = $

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