Question

why is the partial derivative of y^1/3 --- 1/3(y^2/3) and not 1/3(y)^-2/3 ??

Answer 1

i meant and not (1/3)*y^ -2/3

Answer 2

It's not.
\[ \frac{\partial \ }{\partial y} y^{1/3} = \frac{1}{3}y^{-2/3} \]

Answer 3

same thing

Answer 4

Oh, I see. It wasn't clear in your first expression, but there the y is in the denominator of the fraction. Then yes, phi, is right, because
\[ y^{-2/3} = \frac{1}{y^{2/3}} \]

Answer 5

ohhhh. im doing the existence and uniqueness theorem. so i get it. nothing with a negative exponent is continuos at y=0. even if its on the top, its on the bottom.

Answer 6

Right.