Question

If x^2/3 + y^2/3=k ,then dy/dx is equal to

(1 Point)

X/y

(-x/y)^1/3

0

None of these​

Answer 1

${x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = k \\ differentiate \: \: w.r.t \: \: \: x \\ \frac{2}{3} {x}^{ - \frac{1}{3} } + \frac{2}{3} {y}^{ - \frac{1}{3} } \frac{dy}{dx} = 0 \\ \implies {y}^{ - \frac{1}{3} } \frac{dy}{dx} = - {x}^{ - \frac{1}{3} } \\ \implies \frac{dy}{dx} = - \frac{ {x}^{ - \frac{1}{3} } }{ {y}^{ - \frac{1}{3} } } \\ \implies \frac{dy}{dx} = - \frac{ {y}^{ \frac{1}{3} } }{ {x}^{ \frac{1}{3} } } = - ( { \frac{y}{x} })^{ \frac{1}{3} } \\ \implies \frac{dy}{dx} = - \sqrt[3]{ \frac{y}{x} }$