z = x^0.3 y^0.5
I. partial derivative of z with respect to x :
$$\dfrac{\partial z}{\partial x} =0.3x^{-0.7}y^{0.5}$$
II. partial derivative of z with respect to y :
$$\dfrac{\partial z}{\partial y}=0.5x^{0.3}y^{-0.5}$$
Differentiate again to find second - order
partial derivatives:
We can also differentiate $$\small{\frac{ \partial z } { \partial x } \end{array}$$ with respect to y, to find out how it changes when y increases.
This is written as $$\small{\frac{\partial^2 z}{\partial x\ \partial y}}\end{array}$$ and is called a cross-partial derivative:
III. cross partial differential
$$\boxed{\ \dfrac{\partial^2 z}{\partial y\ \partial x}
=\dfrac{\partial^2 z}{\partial x\ \partial y} \ }$$
$$\dfrac{\partial z}{\partial x\ \partial y}=
0.15\ \cdot x^{-0.7}y^{-0.5}$$
I don't know what a cross partial differential is but i think I can answer the partial differentials
Could another mathematician finish it and check what I have done please ?
I am not sure it I am even using the correct symbols.
$$\\z = x^{0.3} y^{0.5}\\\\
\frac{\delta z}{\delta x}=0.3x^{-0.7}y^{0.5}\\\\
\frac{\delta z}{\delta y}=x^{0.3}*0.5y^{-0.5}=0.5x^{0.3}y^{-0.5}\\\\\\
$The rest is for my benefit, I'd like another mathematician to check it please.$\\\\
z = x^{0.3} y^{0.5}\\\\
\frac{dz}{dx}=0.3x^{-0.7}y^{0.5}+x^{0.3}*0.5*y^{-0.5}\;\frac{\delta y}{\delta x}\\\\
\frac{dz}{d y}=0.5x^{0.3}y^{-0.5}+y^{0.5}*0.3x^-0.7\;\frac{\delta x}{\delta y}\\\\\\\\\\$$
z = x^0.3 y^0.5
I. partial derivative of z with respect to x :
$$\dfrac{\partial z}{\partial x} =0.3x^{-0.7}y^{0.5}$$
II. partial derivative of z with respect to y :
$$\dfrac{\partial z}{\partial y}=0.5x^{0.3}y^{-0.5}$$
Differentiate again to find second - order
partial derivatives:
We can also differentiate $$\small{\frac{ \partial z } { \partial x } \end{array}$$ with respect to y, to find out how it changes when y increases.
This is written as $$\small{\frac{\partial^2 z}{\partial x\ \partial y}}\end{array}$$ and is called a cross-partial derivative:
III. cross partial differential
$$\boxed{\ \dfrac{\partial^2 z}{\partial y\ \partial x}
=\dfrac{\partial^2 z}{\partial x\ \partial y} \ }$$
$$\dfrac{\partial z}{\partial x\ \partial y}=
0.15\ \cdot x^{-0.7}y^{-0.5}$$
The cross-partial derivatives are defined as:
zxy = ∂/∂y(∂z/∂x) and zyx = ∂/∂x(∂z/∂y)
so, with z = x0.3y0.5
First do a partial derivative with respect to x: zx = 0.3x-0.7y0.5
Now do a partial derivative with respect to y on the result: zxy = 0.15x-0.7y-0.5
Now try zyx yourself. (Ah, heureka's given the game away!)
.
It should be pointed out that the "cross partial derivatives" are not always equal, but usually they are !
Thankyou Alan, Heureka and Bertie,
I have only just found the time to get back to this question.
Finding the cross partial derivatives is easy thank you for showing me :)
I am trying to think about what these mean.
A normal first derivative gives a gradient of a tangent.
What do partial derivatives and cross partial derivatives represent?
Can anyone give me a relatively simple maybe a pictorial presentation for these?
Maybe it is more complicated that that?
Thank you