Finding 1st and 2nd partial derivitives

**LooNiE** · March 2nd 2009, 07:51 AM

The starting function I have is f=(1/6)x^3 + 2xy^2 + xy
I need to find the 1st and 2nd partial derivatives, and hence find the location and nture of the four stationary points. How do I go about doing that?

**running-gag** · March 2nd 2009, 08:29 AM

Hi

To find the partial derivative with respect to x of $f(x) = \frac{1}{6}\:x^3 + 2xy^2 + xy$ you must consider y as constant

$\frac{\partial f}{\partial x} = \frac{1}{6}\:3\:x^2 + 2y^2 + y = \frac{1}{2}\:x^2 + 2y^2 + y$

Same method to calculate $\frac{\partial f}{\partial y}$ : consider x as constant

**Thomas154321** · March 2nd 2009, 08:31 AM

$f=\frac{1}{6}x^3+2xy^2+xy$

Partial differentiation basically involves choosing a variable to differentiate with respect to, and treating all others as constants. So:

$\frac{\partial f}{\partial x}=f_x=\frac{1}{2}x^2+2y^2+y$

$\frac{\partial f}{\partial y}=f_y=4x+x=5x$

I leave you to find

$\frac{\partial ^2f}{\partial x^2}=f_{xx}$
$\frac{\partial ^2f}{\partial x\partial y}=f_{xy}=f_{yx}=\frac{\partial ^2f}{\partial y\partial x}$
$\frac{\partial ^2f}{\partial y^2}=f_{yy}$

**running-gag** · March 2nd 2009, 08:37 AM

Originally Posted by Thomas154321

$\frac{\partial f}{\partial y}=f_y=4x+x=5x$

Sorry but there is a little mistake

$\frac{\partial f}{\partial y}=f_y=4xy+x$

**LooNiE** · March 2nd 2009, 08:43 AM

Ok I see where you're coming from, as I basically did the same thing before I made this post. I just needed to see if my working as correct. The only problem now for me was to find the stationary points and their nature, i.e. max or min. This is where I am stuck, the first part was pretty easy.

Originally Posted by Thomas154321

$f=\frac{1}{6}x^3+2xy^2+xy$

Partial differentiation basically involves choosing a variable to differentiate with respect to, and treating all others as constants. So:

$\frac{\partial f}{\partial x}=f_x=\frac{1}{2}x^2+2y^2+y$

$\frac{\partial f}{\partial y}=f_y=4x+x=5x$

I leave you to find

$\frac{\partial ^2f}{\partial x^2}=f_{xx}$
$\frac{\partial ^2f}{\partial x\partial y}=f_{xy}=f_{yx}=\frac{\partial ^2f}{\partial y\partial x}$
$\frac{\partial ^2f}{\partial y^2}=f_{yy}$

**Soroban** · March 2nd 2009, 12:59 PM

Hello, LooNiE!

The starting function is: . $f(x,y)\:=\:\tfrac{1}{6}x^3 + 2xy^2 + xy$
I need to find the 1st and 2nd partial derivatives,
and hence find the location and nature of the four stationary points.
How do I go about doing that? . . And no one has told you how?

To find the stationary points, set the first partials equal to zero and solve.

. . . $\begin{array}{ccccccc}f_x &=&\frac{1}{2}x^2 + 2y^2 + y &=&0 & {\color{red}[1]} \\ \\[-3mm]<br /> f_y &=&4xy + x &=&0 & {\color{red}[2]}\end{array}$

From [2], we have: . $x(4y+1) \:=\:0 \quad\Rightarrow\quad x = 0,\:y = \text{-}\tfrac{1}{4}$

Substitute $x=0$ into [1]: . $2y^2 + y \:=\:0 \quad\Rightarrow\quad y \:=\:0,\:\text{-}\tfrac{1}{2}$
. . We have two of the points: . ${\color{blue}(0,0),\:\left(0,\text{-}\tfrac{1}{2}\right)}$

Substitute $y = \text{-}\tfrac{1}{4}$ into [1]: . $\tfrac{1}{2}x^2 + \tfrac{1}{8} -\tfrac{1}{4} \:=\:0 \quad\Rightarrow\quad x = \pm\tfrac{1}{2}$
. . And we have two more points: . ${\color{blue}\left(\tfrac{1}{2},\:\text{-}\tfrac{1}{4}\right),\:\left(\text{-}\tfrac{1}{2},\:\text{-}\tfrac{1}{4}\right)}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

For the Second Partial Test, form the expression: . $D \;=\;\left(f_{xx}\right)\left(f_{yy}\right) - \left(f_{xy}\right)^2$

We have: . $f_{xx}\:=\:x,\;\;f_{yy}\:=\:4x,\;\;f_{xy} \:=\:4y+1$

. . Then: . $D \;=\;4x^2 - (4y+1)^2$

At ${\color{blue}(0,0)}\!:\; D\:=\:4(0) - (0+1)^2 \:=\:-1$ . . . negative: saddle point

At ${\color{blue}\left(0,\text{-}\tfrac{1}{2}\right)}\!:\;D \:=\:4(0) - (\text{-}2+1)^2 \:=\:-1$ . . . negative: saddle point

At ${\color{blue}\left(\tfrac{1}{2},\text{-}\tfrac{1}{4}\right)}\!:\;D\:=\:4(\tfrac{1}{4}) - (-1+1)^2 \:=\:+1$
. . . $f_{xx} \:=\:+\tfrac{1}{2}$ . . . positive, concave up, $\cup$ . . . minimum

At ${\color{blue}\left(\text{-}\tfrac{1}{2},\text{-}\tfrac{1}{4}\right)}\!:\;D \:=\:4(\tfrac{1}{4}) - (-1+1)^2 \:=\:+1$
. . . $f_{xx} \:=\:\text{-}\tfrac{1}{2}$ . . . negative, concave down, $\cap$ . . . maximum

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