If {x^{2/3}} + {y^{2/3}} = {a^{2/3}},then fra | vedclass

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(B) Given the equation: ${x^{2/3}} + {y^{2/3}} = {a^{2/3}}$Differentiating both sides with respect to $x$:$\frac{d}{dx}({x^{2/3}}) + \frac{d}{dx}({y^{

Vedclass · Accepted Answer

$-\left( \frac{y}{x} \right)^{1/3}$

Vedclass · Answer

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

If ${x^{2/3}} + {y^{2/3}} = {a^{2/3}}$,then $\frac{dy}{dx} = $

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