Finding Partial Derivatives of Implicit Functions

[arl146]

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 can't get dz/dx on one side by itself... help please

 

[I like Serena]

Hi arl146!

Suppose your equation was: x = A * (1 + x)
Could you solve it then?

 

[arl146]

Am I still doing that in terms of a partial derivative? I don't understand your example

 

[I like Serena]

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.

 

[arl146]

Ohh ok. Embarrassing but no I can't haha after I get x/(1+x) idk what to do next

 

[I like Serena]

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?

 

[arl146]

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

 

[I like Serena]

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?

 

[arl146]

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))

 

[arl146]

That's prob not right...

 

[I like Serena]

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:
z = \sin(x+y+z)
{\partial z \over \partial x} = \cos(x+y+z) {\partial \over \partial x}(x+y+z)
{\partial z \over \partial x} = \cos(x+y+z) (1 + {\partial z \over \partial x})

Now you need to solve for {\partial z \over \partial x}.

 

[arl146]

Oh I got it now. Sorry I'm dumb lol
dz/dx= cos(x+y+z) / (1+cos(x+y+z))

 

[Mark44]

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))

 

[I like Serena]

Oh I got it now. Sorry I'm dumb lol
dz/dx= cos(x+y+z) / (1+cos(x+y+z))

Almost!

Could you check your plus and minus signs?

 

[arl146]

oh yea, i had 1-cos(stuff) not +. again, was on my phone earlier. and thanks!

 

[I like Serena]

You're welcome!