Question

If the following is true, find fx(4, 0).
f(x,y) = root3(x^3+y^3)

Answer 1

Fill in x,y... x=4 y=0 right?

Answer 2

This is calc 3, do you know this?

Answer 3

It's actually f sub x (4, 0)

Answer 4

Alright no need for attitude. Have a nice time figuring it out.

Answer 5

The way you wrote it is a little strange.
root3?

Did you mean to say `cube root` or is the 3 underneath a `square root` ?

Answer 6

\[\Large f(x,y)\quad=\quad (x^3+y^3)^{1/3}\]So we differentiate the outer function, then the chain rule tells us to multiply by the derivative of the inner function.\[\Large f_x(x,y)\quad=\quad \frac{1}{3}(x^3+y^3)^{-2/3}\cdot \frac{\partial}{\partial x}(x^3+y^3)\]

Answer 7

Since we're taking a partial derivative, with respect to x, we treat this y^3 as constant.
The derivative of a constant is 0.\[\Large f_x(x,y)\quad=\quad \frac{1}{3}(x^3+y^3)^{-2/3}\cdot (3x^2+0)\]

Answer 8

\[\Large f_x(x,y)=\frac{3x^2}{3(x^3+y^3)^{2/3}}\]

Answer 9

\[\Large f_x(4,0)=\frac{3\cdot 4^2}{3(4^3+0^3)^{2/3}}\]

Answer 10

Hopefully that was what you meant by root3 :\