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# What is the cross-partial derivatives for z = x^0.3 y^0.5 ???

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What is the cross-partial derivatives for z = x^0.3 y^0.5 ???

Mar 29, 2015

#2
+20850
+11

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

$$\boxed{\ \dfrac{\partial^2 z}{\partial y\ \partial x} =\dfrac{\partial^2 z}{\partial x\ \partial y} \ }$$

$$\dfrac{\partial z}{\partial x\ \partial y}= 0.15\ \cdot x^{-0.7}y^{-0.5}$$

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Mar 29, 2015

#1
+95360
0

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.

$$\\z = x^{0.3} y^{0.5}\\\\ \frac{\delta z}{\delta x}=0.3x^{-0.7}y^{0.5}\\\\ \frac{\delta z}{\delta y}=x^{0.3}*0.5y^{-0.5}=0.5x^{0.3}y^{-0.5}\\\\\\ The rest is for my benefit, I'd like another mathematician to check it please.\\\\ z = x^{0.3} y^{0.5}\\\\ \frac{dz}{dx}=0.3x^{-0.7}y^{0.5}+x^{0.3}*0.5*y^{-0.5}\;\frac{\delta y}{\delta x}\\\\ \frac{dz}{d y}=0.5x^{0.3}y^{-0.5}+y^{0.5}*0.3x^-0.7\;\frac{\delta x}{\delta y}\\\\\\\\\\$$

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Mar 29, 2015
#2
+20850
+11

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

$$\boxed{\ \dfrac{\partial^2 z}{\partial y\ \partial x} =\dfrac{\partial^2 z}{\partial x\ \partial y} \ }$$

$$\dfrac{\partial z}{\partial x\ \partial y}= 0.15\ \cdot x^{-0.7}y^{-0.5}$$

heureka Mar 29, 2015
#3
+27377
+8

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!)

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Mar 29, 2015
#4
+20850
+3

Alan, sorry

Mar 29, 2015
#5
+890
+5

It should be pointed out that the "cross partial derivatives" are not always equal, but usually they are !

Mar 29, 2015
#6
+95360
0

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

Mar 31, 2015