What is the derivative of #x^(2/3)+y^(2/3)=5# at the given point of #(8,1)#?

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Answer 1 · 2016-10-17T06:55:51

#dy/dx = -1/2# at #(x,y) = (8, 1)#

First, let's find #dy/dx# using implicit differentiation:

#d/dx(x^(2/3)+y^(2/3))=d/dx5#

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

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

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

Now, we evaluate #dy/dx# at our given point of #(x,y) = (8,1)#

#dy/dx|_((x,y)=(8,1)) = -(8/1)^(-1/3)#

#=-8^(-1/3)#

#=-1/2#