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Find partial derivative dz/dx

  • Thread starter arl146
  • Start date
  • #1
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Homework Statement


Consider z=sin(x+y+z). This defines z implicitly as a function of x and y. Find an expression for dz/dx



The Attempt at a Solution


This was on a test, this is what i did. I got 7/11 pts

dz/dx = cos(x+y+z)*(1+(dz/dx))

(dz/dx) / (1 + (dz/dx)) = cos(x+y+z)

i cant get dz/dx on one side by itself.... help please
 

Answers and Replies

  • #2
I like Serena
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Hi arl146! :smile:

Suppose your equation was: x = A * (1 + x)
Could you solve it then?
 
  • #3
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Am I still doing that in terms of a partial derivative? I don't understand your example
 
  • #4
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Am I still doing that in terms of a partial derivative? I don't understand your example
I've replaced the complex sub expressions in your equation by A and x respectively.
The result is a simpler equation.
If you can solve x from that, you should also be able to find dz/dx equivalently.
 
  • #5
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Ohh ok. Embarrassing but no I can't haha after I get x/(1+x) idk what to do next
 
  • #6
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The method to solve an equation like this is:
x = A * (1 + x)
Remove parentheses: x = A * 1 + A * x
Subtract A*x on both sides: x - A * x = A * 1
Extract x, introducing parentheses again: x * (1 - A) = A * 1
Divide both side by (1-A): x = (A * 1) / (1 - A)

Does this look familiar?
 
  • #7
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Ohh yea ... Don't know why I couldn't do that.. So for my problem, would I get all the z on one side? If I do that I get inverse sin(z)- z=x+y

Sorry for the lack of normal format, I'm on my phone
 
  • #8
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Hmm, how did you get that?

It is not right.
You have a cosine. How did you derive an inverse sine from that?
And actually, you should not bring all z to one side, but only all (dz/dx).

Can you perhaps show the steps you took?
 
  • #9
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Oh I didn't do the derivative if that yet. But I just did it another way while waiting for you to respond:

Z=sin(x+y+z)
z=sin(x)+sin(y)+sin(z)
z-sin(z)=what's left
(dz/dx)-cos(z)*(dz/dx)=same stuff
(dz/dx)(1-cos(z))=sin(x+y)
dz/dx= (sin(x+y))/(1-cos(z))
 
  • #10
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That's prob not right...
 
  • #11
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Oh I didn't do the derivative if that yet. But I just did it another way while waiting for you to respond:

Z=sin(x+y+z)
z=sin(x)+sin(y)+sin(z)
z-sin(z)=what's left
(dz/dx)-cos(z)*(dz/dx)=same stuff
(dz/dx)(1-cos(z))=sin(x+y)
dz/dx= (sin(x+y))/(1-cos(z))
That's prob not right...
Sorry, but I'm afraid you can't do:
Z=sin(x+y+z)
z=sin(x)+sin(y)+sin(z)

Your original first step was good.
That is:
[itex]z = \sin(x+y+z)[/itex]
[itex]{\partial z \over \partial x} = \cos(x+y+z) {\partial \over \partial x}(x+y+z)[/itex]
[itex]{\partial z \over \partial x} = \cos(x+y+z) (1 + {\partial z \over \partial x})[/itex]

Now you need to solve for [itex]{\partial z \over \partial x}[/itex].
 
  • #12
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Oh I got it now. Sorry I'm dumb lol
dz/dx= cos(x+y+z) / (1+cos(x+y+z))
 
  • #13
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Oh I didn't do the derivative if that yet. But I just did it another way while waiting for you to respond:

Z=sin(x+y+z)
z=sin(x)+sin(y)+sin(z)
That's definitely not right. sin(x + y + z) ≠ sin(x) + sin(y) + sin(x)
z-sin(z)=what's left
(dz/dx)-cos(z)*(dz/dx)=same stuff
(dz/dx)(1-cos(z))=sin(x+y)
dz/dx= (sin(x+y))/(1-cos(z))
 
  • #14
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Oh I got it now. Sorry I'm dumb lol
dz/dx= cos(x+y+z) / (1+cos(x+y+z))
Almost! :wink:

Could you check your plus and minus signs?
 
  • #15
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oh yea, i had 1-cos(stuff) not +. again, was on my phone earlier. and thanks!
 
  • #16
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You're welcome! :smile:
 

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