Question

What's the partial derivative w=rad(25-5x^2-5y^2)?

The equation's longer, but I just need this piece of info cause I can't figure out the partial derivative of a radical

Answer 1

use the chain rule just like in calc I, and treat the variables you are not differentiating with respect to as constants

Answer 2

Does it stay in the radical, or does it become a fraction with the original equation in the denominator and the partial derivative in the numerator?

Answer 3

let \(u=25-5x^2-5y^2\implies w=\sqrt u\), then
\[\frac{\partial w}{\partial x}=\frac12u^{-1/2}\cdot\frac{\partial u}{\partial x}\]is that enough info?

Answer 4

As @TuringTest said, let u = inside bit, then w' = u'/(2w)

Answer 5

So.....

\[\frac{ -10x }{ 2\sqrt{25-5x ^{2}-5y ^{2}} }\]?

Answer 6

that's what I got, but you can simplify -10/2

Answer 7

Ah, yes you can.... Alrighty, thanks!