Partial differention hard questions

In summary: I THINK I DID IT WRONG HAVE NO IDEAisnt der of z wrt y of sqrt(x) = 0isnt der of z wrt y of sqrt(x) = 0
  • #1
dagg3r
67
0
HI GUYS
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 that's 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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  • #2
isnt der of z wrt y of sqrt(x) = 0
 
  • #3
mathmike said:
isnt der of z wrt y of sqrt(x) = 0
Yes, it is that's why the derivative of [tex]4y^3\sqrt{x}[/tex] is
[tex]12y^2\sqrt{x}[/tex] just as dagg3r says- even if you use the "product rule". What did you think the derivative should be?
The partial derivative of [tex]z= 4y^3\sqrt{x} - ln(4xy^3) + 2x[/tex]
with respect to y is:
[tex]\frac{\partial z}{\partial y}= 12y^2\sqrt{x}- \frac{1}{4xy^3}\left(12xy^2)\right)[/tex]
= [tex]12y^2\sqrt{x}- 3y[/tex] exactly what dagg3r had!
(Click on the TEX above to see the code I used- your "day" is driving me mad! I would have preferred you just use "d" with the notation that it is partial differentiation.)
dagg3r said:
2. FInd the stationary point of
z= 5x^2 + 23x + y^2 + 8y + 3xy + 32 amd determine the nature of this point
Okay, you got
[tex]\frac{\partial z}{\partial x}= 10x+23+ 3y[/tex]
and
[tex]\frac{\partial z}{\partial y}= 2y+ 8+ 3x[/tex]
Of course, at a stationary point, the partial derivatives are 0:
10x+ 3y+23= 0 and
3x+ 2y+ 8= 0 (You did not write "= 0". While I understood what you were doing, the "= 0" makes it clearer. Also I have written the two equations with x first, then y. That also makes it a little simpler to see what to do.
Yes, multiply the first equation by 2 and the second by 3, to get "6y" in both and then subtract:
20x+ 6y+ 46= 0
9x+ 6y+ 24= 0
which gives
11x+ 22= 0.
You subtracted wrong! (That might be because of the way you had written the equations.)
x= -2, not 2, and then, from the second initial equation, -6+ 2y+ 8= 0 so
2y= -8+ 6= -2, y= -1. The stationary point is at (2, -1).
z(2, -1)= 5(4) + 23(2) + (-1)^2 + 8(-1) + 3(2)(-1) + 32
= 20+ 46+ 1- 8- 6+ 32= 85.
NOW determine the nature of the stationary point.
 

1. What is partial differentiation?

Partial differentiation is a mathematical process used to find the rate of change of a multivariable function with respect to one of its variables, while holding all other variables constant.

2. Why are partial differentiation hard questions?

Partial differentiation can be challenging because it involves manipulating multiple variables and understanding how they are related to each other. It also requires a strong understanding of calculus and algebraic techniques.

3. How is partial differentiation different from ordinary differentiation?

Ordinary differentiation involves finding the derivative of a function with respect to a single variable, while partial differentiation involves finding the derivative with respect to one variable while holding all other variables constant.

4. What are some real-world applications of partial differentiation?

Partial differentiation is used in many fields, including physics, economics, and engineering. It can be used to optimize functions, analyze the behavior of systems, and solve problems involving multiple variables.

5. How can I improve my skills in solving partial differentiation hard questions?

Practice is key in improving your skills in partial differentiation. Familiarize yourself with the basic rules and techniques, and then try solving a variety of problems to build your confidence and understanding. You can also seek out resources such as textbooks, online tutorials, and practice problems to help you improve.

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