Question

Calculate the second order partial derivatives of f(x,y) = sqrt(x+y^5)

Answer 1

With respect to which variable?

Answer 2

Both x and y if that's what you meant.

Answer 3

df/dx =5x^4^(1/2)

dx'' = 20x^3^(-1/2)

df/dy = 5y^4^(1/2)

dy'' = 20y^3^(-1/2)

Answer 4

Hint:\[\frac{\partial}{\partial x}\left [ \sqrt{x+y^5} \right ]=\frac{1}{2\sqrt{x+y^5}},\]\[\frac{\partial^2}{\partial x^2}\left [ \sqrt{x+y^5} \right ]=\frac{\partial}{\partial x}\left [ \frac{1}{2\sqrt{x+y^5}} \right ]=-\frac{1}{4(x+y^5)^\frac{3}{2}}.\]

Answer 5

That's for x. Can you do it for y?

Answer 6

Ahhh, I get it now. I think I can do y now. Thanks!

Answer 7

and \[\frac{\partial^2}{\partial x\partial y}\]

Answer 8

Oh right I forgot about that lol. How do I find the xy for it?

Answer 9

\[\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}(x+y^5)^{1/2}\right)=\frac{\partial}{\partial y}\left(\frac{1}{2}(x+y^5)^{-1/2}\right)=\]\[=\frac{1}{2}\cdot\frac{-1}{2}(x+y^5)^{-3/2}\cdot 5y^4=-\frac{5}{4}y^4(x+y^5)^{-3/2}\]