#2**+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**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**+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**+8 **

The cross-partial derivatives are defined as:

z_{xy} = ∂/∂y(∂z/∂x) and z_{yx} = ∂/∂x(∂z/∂y)

so, with z = x^{0.3}y^{0.5}

First do a partial derivative with respect to x: z_{x} = 0.3x^{-0.7}y^{0.5}

Now do a partial derivative with respect to y on the result: z_{xy} = 0.15x^{-0.7}y^{-0.5}

Now try z_{yx} yourself. (Ah, heureka's given the game away!)

.

Alan
Mar 29, 2015

#5**+5 **

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

Bertie
Mar 29, 2015

#6**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