Is This Calculation of ∂z/∂x Correct for the Given Function?

njo
Messages
20
Reaction score
0

Homework Statement


∂z/∂x of ycos(xz)+(4xy)-2z^2x^3=5x[/B]

Homework Equations


n/a

The Attempt at a Solution


∂z/∂x=(5+yz-4y+6z^2x^2)/(-yxsin(xz)-4zx^3)[/B]

Is this correct? Just trying to make sure that's the correct answer. I appreciate the help. I can post my work if need be. Thanks
 
Physics news on Phys.org
Close, but check your work. There is at least a sine term missing in the numerator. Better yet, show your work.
 
-y*sin(xz)*(z+x(∂z/∂x))+4y-4zx^3(∂z/∂x)-6z^2x^2 = 5

This is what I have before rearranging and factoring for ∂z/∂x
 
njo said:
-y*sin(xz)*(z+x(∂z/∂x))+4y-4zx^3(∂z/∂x)-6z^2x^2 = 5

This is what I have before rearranging and factoring for ∂z/∂x

That looks good. If you carefully do the algebra solving for ##\frac{\partial z}{\partial x}## you should be OK.
 
(5+yzsin(xz)-4y+6z^2x^2)/(-yxsin(xz)-4zx^3) = ∂z/∂x

Pretty sure this is right. Just messed up on my algebra. Thank you so much. The internet is great.
 
Thread 'Use greedy vertex coloring algorithm to prove the upper bound of χ'
Hi! I am struggling with the exercise I mentioned under "Homework statement". The exercise is about a specific "greedy vertex coloring algorithm". One definition (which matches what my book uses) can be found here: https://people.cs.uchicago.edu/~laci/HANDOUTS/greedycoloring.pdf Here is also a screenshot of the relevant parts of the linked PDF, i.e. the def. of the algorithm: Sadly I don't have much to show as far as a solution attempt goes, as I am stuck on how to proceed. I thought...
Back
Top