HI GUYS(adsbygoogle = window.adsbygoogle || []).push({});

i just have some hard problems on partial differentiation... hope you guys can help out

1. find "day z / day y" its the funny backwards 6 symbols instead of the normal differntiation symbole

if z=4y^3 * sqrt(x) - ln(4xy^3) + 2x

i did this:

i diffenriated 4y^3 * sqrt(x) with respect to y, and since sqrt(x) stays constant, it remains hence i get

12y^2*sqrt(x) . i then diffed ln(4xy^3) + 2x

i did

ln(u)

u=4xy^3 u`= 12xy^2

=1/u * u`

=(1/4xy^3) * 12xy^2

= 3/y

and the 2x disappears

hence my ultimate answer is

"day z/ day y"=12y^2*sqrt(x) - 3/y

2. FInd the stationary point of

z= 5x^2 + 23x + y^2 + 8y + 3xy + 32 amd determine the nature of this point

day z / day x = 10x + 23 + 3y

day z / day y = 2y + 8 + 3x

use elimination hence

10x + 23 + 3y *2

2y + 8 + 3x * 3

-----------------

20x + 46 + 6y 3)

6y + 24 + 9x 4)

---------------

11x -22 = 0

x = 2 sub this into eqn 3)

y= -43/3

now sub x=2, y=-43/3 into z=

hence z=5(2)^2 + 23(2) + (-43/3)^2 + 8(-43/3) + 3*(2)(-43/3) + 32

z=925/9

hence stationary points i think are (2, -43/3 , 925/9)

ok assuming thats right i have to determine the nature of the point, to my knowledge i would have to

use the formula g= (day^2 z/day*x^2)*(day^2 z/ day*y^2) - ( day^2*z / day*x*day*y )

day^2 z / day^2 x^2 = 10

day^2 z / day^2 y^2 = 2

day^2*z / day*x*day*y = 3

hence g = (10)*(2) - 3

G= 17

HENCE 17 > 0 , AS (day^2 z / day^2 x^2) IS > 0, AND G> 0, THEN THE STATIONARY POINT IS A MINUMUM POINT

I THINK I DID IT WRONG HAVE NO IDEA

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# Homework Help: Partial differention hard questions

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