Let x^{k}+y^{k}=a^{k} where a, k 0. If frac | vedclass

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(C) Given the equation $x^{k}+y^{k}=a^{k}$.Differentiating both sides with respect to $x$,we get:$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$.Dividing by

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$\frac{2}{3}$

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(C) Given the equation $x^{k}+y^{k}=a^{k}$.Differentiating both sides with respect to $x$,we get:$k x^{k-1} + k y^{k-1} \frac{dy}{dx} = 0$.Dividing by $k$ (since $k > 0$):$x^{k-1} + y^{k-1} \frac{dy}{dx} = 0$.Rearranging to solve for $\frac{dy}{dx}$:$\frac{dy}{dx} = -\frac{x^{k-1}}{y^{k-1}} = -\left(\frac{x}{y}\right)^{k-1}$.We are given $\frac{dy}{dx} + \left(\frac{y}{x}\right)^{\frac{1}{3}} = 0$,which implies $\frac{dy}{dx} = -\left(\frac{y}{x}\right)^{\frac{1}{3}} = -\left(\frac{x}{y}\right)^{-\frac{1}{3}}$.Comparing the two expressions for $\frac{dy}{dx}$:$-\left(\frac{x}{y}\right)^{k-1} = -\left(\frac{x}{y}\right)^{-\frac{1}{3}}$.Equating the exponents:$k - 1 = -\frac{1}{3}$.Solving for $k$:$k = 1 - \frac{1}{3} = \frac{2}{3}$.

Let $x^{k}+y^{k}=a^{k}$ where $a, k > 0$. If $\frac{dy}{dx}+\left(\frac{y}{x}\right)^{\frac{1}{3}}=0$,then the value of $k$ is:

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