# Find the second partials F(s,t)=sqrt s^2 + t^2Including the mix partials

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### 1 Answer

I am going to assume you mean

`sqrt(s^2 + t^2)` , otherwise it seems a little odd to ask about the mixed partials.

`F_s` means treat `s` like a variable and everything else (here, `t`) like a constant, and take the derivative:

`F(s,t) = (s^2 + t^2)^(1/2)`

`F_s = (1/2) (s^2 + t^2)^(-1/2) * 2s = s(s^2 + t^2)^(-1/2)`

`F_t = (1/2) (s^2 + t^2)^(-1/2) * 2t = t(s^2 + t^2)^(-1/2)`

`F_(ss)` means start with `F_s` and again treat `s` like the variable. To do that, we need the product rule:

`F_(ss) = 1*(s^2 + t^2)^(-1/2) + s * (-1/2) (s^2 + t^2)^(-3/2) * 2s`

`(s^2 + t^2)^(-1/2) - s^2 (s^2 + t^2)^(-3/2)`

`F_(tt) = 1*(s^2 + t^2)^(-1/2) + t * (-1/2) (s^2 + t^2)^(-3/2) * 2t`

`(s^2 + t^2)^(-1/2) - t^2 (s^2 + t^2)^(-3/2)`

`F_(st)` means start with `F_s` and now treat `t` like the variable, and `s` like the constant:

`F_s=s(s^2 + t^2)^(-1/2)`

That `s` out front is now just a "constant" so we don't need the product rule:

`F_(st)=s * (-1/2) (s^2 + t^2)^(-3/2) * 2t = -st (s^2 + t^2)^(-3/2)`

Or we could have started with `F_t` and taken the derivative with `s` as the variable and gotten the same thing:

`F_(ts) = t * (-1/2) (s^2 + t^2)^(-3/2) * 2s = -st (s^2 + t^2)^(-3/2)`