Question

if x^2/3 + y^2/3 =2 ,find dy/dx at (1,1)

Answer 1

Answer:

Step-by-step explanation:

x^2/3 +y^2/3=2

differentiating both sides w.r.t  x

d/dx(x^2/3 +y^2/3)=d/dx(2)

2/3 x^-1/3 + (2/3 y^-1/3) dy/dx=0

(2/3 y^-1/3) dy/dx= -2/3 x^-1/3

dy/dx=(-2/3 x^-1/3)/(2/3 y^-1/3)

dy/dx=  (- y^1/3 ) / ( x^1/3)

dy/dx (1,1)= -(1)^1/3/ 1()^1/3

dy/dx(1,1)=  -1      (answer)

Answer 2

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