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

difficulty advanced
 Mar 29, 2015

Best Answer 

 #2
avatar+26396 
+11

 z = x^0.3 y^0.5 

I. partial derivative of z with respect to x :

zx=0.3x0.7y0.5

 

II. partial derivative of z with respect to y :

zy=0.5x0.3y0.5

 

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

 2zy x=2zx y 

zx y=0.15 x0.7y0.5

 Mar 29, 2015
 #1
avatar+118703 
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=x0.3y0.5δzδx=0.3x0.7y0.5δzδy=x0.30.5y0.5=0.5x0.3y0.5$Therestisformybenefit,Idlikeanothermathematiciantocheckitplease.$z=x0.3y0.5dzdx=0.3x0.7y0.5+x0.30.5y0.5δyδxdzdy=0.5x0.3y0.5+y0.50.3x0.7δxδy

 Mar 29, 2015
 #2
avatar+26396 
+11
Best Answer

 z = x^0.3 y^0.5 

I. partial derivative of z with respect to x :

zx=0.3x0.7y0.5

 

II. partial derivative of z with respect to y :

zy=0.5x0.3y0.5

 

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

 2zy x=2zx y 

zx y=0.15 x0.7y0.5

heureka Mar 29, 2015
 #3
avatar+33654 
+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!)

 

.

 Mar 29, 2015
 #4
avatar+26396 
+3

Alan, sorry

 Mar 29, 2015
 #5
avatar+893 
+5

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

 Mar 29, 2015
 #6
avatar+118703 
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

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