Let x k + y k = a k , a , k > 0 and d y d x + y x 1 3 = 0 , then k is

Differentiating the given equation with respect to x k . x k - 1 + k . y k - 1 d y d x = 0 ⇒ d y d x = - x y k - 1 ⇒ d y d x + x y k - 1 = 0 ⇒ k - 1 = - 1 3 ⇒ k = 1 - 1 3 = 2 3

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Mathematics - Differentiation Question with Solution | TestHub

MathematicsDifferentiationDifferentiation of implicit functionsMedium2 minPYQ_2020

MathematicsMediumsingle choice

Let $x^{k} + y^{k} = a^{k}, (a, k > 0)$ and $\frac{d y}{d x} + {(\frac{y}{x})}^{\frac{1}{3}} = 0,$ then $k$ is

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Solution:

Differentiating the given equation with respect to $x$

$k . x^{k - 1} + k . y^{k - 1} \frac{d y}{d x} = 0$

$\Rightarrow \frac{d y}{d x} = - {(\frac{x}{y})}^{k - 1}$

$\Rightarrow \frac{d y}{d x} + {(\frac{x}{y})}^{k - 1} = 0$

$\Rightarrow k - 1 = - \frac{1}{3}$

$\Rightarrow k = 1 - \frac{1}{3} = \frac{2}{3}$

Stream:JEESubject:MathematicsTopic:DifferentiationSubtopic:Differentiation of implicit functions

⏱ 2mℹ️ Source: PYQ_2020

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