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

difficulty advanced
Guest Mar 29, 2015

Best Answer 

 #2
avatar+19653 
+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
 #1
avatar+92806 
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.     

 

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

Melody  Mar 29, 2015
 #2
avatar+19653 
+11
Best Answer

 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
avatar+26753 
+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!)

 

.

Alan  Mar 29, 2015
 #4
avatar+19653 
+3

Alan, sorry

heureka  Mar 29, 2015
 #5
avatar+889 
+5

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

Bertie  Mar 29, 2015
 #6
avatar+92806 
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 

Melody  Mar 31, 2015

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