How do you use implicit differentiation to find #(dy)/(dx)# given #x^(2/3)+y^(2/3)=1#?

2 Answers
Aug 29, 2016

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

Explanation:

Use the power rule and implicit differentiation to find the derivative.

#d/dx(x^(2/3) + y^(2/3)) = d/dx(1)#

#d/dx(x^(2/3)) + d/dx(y^(2/3)) = d/dx(1)#

Since we're differentiating with respect to #x#, we must put #dy/dx# at every y term that we differentiate (after differentiating, of course).

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

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

Isolate #dy/dx# and simplify:

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

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

#dy/dx = (-6y^(1/3))/(6x^(1/3))#

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

Hopefully this helps!

Aug 29, 2016

# -y^(1/3)/x^(1/3)#

Explanation:

Define #f(x,y)=x^(2/3)+y^(2/3)-1=0#

we know that

#df = f_x dx + f_y dy = 0#

so

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