Question

find the first order partial derivatives of
f(s,t)=(s-t)/(s+t)

Answer 1

You have\[f(s,t)=\frac{s-t}{s+t}\]and are asked to find both \(f_s(s,t)\) and \(f_t(s,t)\). Here's a hint:\[f_s(s,t)=\frac\partial{\partial s}\left[\frac s{s+t}\right]-t\frac\partial{\partial s}\left[\frac1{s+t}\right].\]

Answer 2

@across, it would be nice to use the quotient rule also , just reduces the number of steps

Answer 3

@across and save space too

Answer 4

I get lost with the s and ts

Answer 5

Following @psi9epsilon's suggestion,\[\frac\partial{\partial s}\left[\frac{s-t}{s+t}\right]=\frac{(s+t)(s-t)_s-(s+t)_s(s-t)}{(s+t)^2}=\frac{2t}{(s+t)^2}.\]Can you do the other?

Answer 6

I expanded the bottom but thats what messed me up

Answer 7

You don't need to.

Answer 8

All you have to do is compute\[\frac\partial{\partial s}\left[\frac{s-t}{s+t}\right]\]by remembering that\[\frac d{dx}\left[\frac{g(x)}{h(x)}\right]=\frac{g'(x)h(x)-g(x)h'(x)}{[h(x)]^2}.\]

Answer 9

\[\frac{ 2s }{ (s+t)^{2} }\]

is that the second part

Answer 10

Let's check it out:\[\frac\partial{\partial t}\left[\frac{s-t}{s+t}\right]=\frac{(s-t)_t(s+t)-(s-t)(s+t)_t}{(s+t)^2}=\frac{-(s+t)-(s+t)}{(s+t)^2}=\frac{-2}{s+t}.\]Nope. :P But I already solved it for you, so gl.