# Homework Help: Chain Rule with Partials & Evaluation Question

1. Nov 3, 2013

### Alex Bard

Hi, I have a test prep question regarding Chain Rule, please see the problem and my attempt below. I believe part A is okay but part B, I'm just confused, seems like there is a part missing from the question, or at least how I'm use to doing it.

1. The problem statement, all variables and given/known data

$A. Let f(x, y) = cos(xy) + ycos(x), where x = u^2 + v and y = u - v^2. Find \frac{∂z}{∂v} when u = 1, v = -1.$

B. Let f(x,y) = ln(x - 3y). Find f(6.9, 2.06)

2. Relevant equations

Chain Rule is stated by $\frac{∂z}{∂x} * \frac{∂x}{∂v} + \frac{∂z}{∂y} * \frac{∂y}{∂v}$

3. The attempt at a solution

Part A.

$frac{∂z}{∂v}$ = cos(xy) + y cos(x)

First take the partial in respect to x and multiply by partial to respect to v
= (y)[-sin(xy) - y sin(x)] (2u + v)

Take partial in respect to y and multiply by partial in respect to v
= (x)[-sin(xy) + cos(x)] (u - 2v)

Add both of them to complete:
= (y)[-sin(xy) - y sin(x)] (2u + v) + (x)[-sin(xy) + cos(x)] (u - 2v)

Now, I substitute the values of x & y with their respective counterparts:
= (u - v^2)[-sin((u^2+v)(u-v^2)) - (u-v^2)(sin(u^2+v))] + (u^2+v)[-sin((u^2+v)(u-v^2)) + cos(u^2+v)](u-2v)

Now I will plug in the respective values for u & v with their respective counterparts:
= (1 - (-1)^2)[-sin((1^2+(-1))(1-(-1)^2)) - (1-(-1)^2)(sin(1^2+(-1)))] + (1^2+(-1))[-sin((1^2+(-1))(1-(-1)^2)) + cos(1^2+(-1))](1-2(-1))

Now to simplify and Solve:
Now I will plug in the respective values for u & v with their respective counterparts:
= (0)[-sin((0)(0)) - (0)(sin(0))] + (0)[-sin((0)(0)) + cos(0)](3)
= 0 + 0 (3)

so the answer I get is 0.

Does that seem right?

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Part B.

Let f(x, y) = ln(x - 3y). Find (6.9, 2.06)

I believe this problem is missing a point but considering this problem as is, I got:

Find f(6.9, 2.06) = ln(x - 3y)
→ (6.9 - 6.18) = 0.72 → ln(0.72) ≈ -0.33

Now zx = $\frac{1}{x - 3y}$
Now zy = $\frac{-3}{x - 3y}$

$\frac{1}{.72} , \frac{-3}{.72}$

Last edited: Nov 3, 2013
2. Nov 3, 2013

### brmath

I'm unsure of your notation, which makes it hard to answer. By $\int(x,y)$ do you mean f(x,y) a function, and not an integral? And do you then mean z = f(x,y)?

3. Nov 3, 2013

### Alex Bard

Yes sir, for some reason seeing the integral sign at first looked like a f to me, as in function. Fixed the notation.

and

4. Nov 3, 2013

### haruspex

You've lost me already. You don't mean f(x,y) on the LHS, right? Some derivative of f? Even then, I don't see how you get the RHS, e.g. how you would get any sin(y) term. Pls fill in the gaps.

5. Nov 3, 2013

### Alex Bard

Apologize for the messiness, I rewrote and did my best to follow the brackets and (), please let me know if it doesn't make sense for me to go back and clean it up even further.

6. Nov 3, 2013

### brmath

Alex, assuming z = f(x,y) did you mean your next line to be dz/dx which would be $f_x(x,y)$? Because you wrote f(x,y) again. And if you did mean it to be $f_x(x,y)$ it isn't computed correctly. There could not be a sin(y) in that derivative. Pretend to begin with that the "y" is a 2 and it may be clearer how to take this derivative.

Until this much is sorted out, we can't go on to the next step.

7. Nov 3, 2013

### Alex Bard

Sir, I believe I've cleaned it up right before your comment. Per haps it looks better now? Let me know if its still messy and I will attempt to break it up. Also, you are correct, i did mess that part up and re-did my math as a result. Thank you for bringing it to my attention.

8. Nov 3, 2013

### haruspex

That gives you a term -y2 sin(x) on the left. Looks like a factorisation error.
And ∂x/∂v is not (2u + v).

9. Nov 3, 2013

### brmath

You are now on the right track. However your $∂x/∂v$ and $∂y/∂v$ are not computed correctly.

10. Nov 3, 2013

### Alex Bard

Okay, I see that it should be (2u + 1) in ∂x/∂v

and I see, it was an ill placed bracket, the y should NOT be distributed across both terms. It should be:
= [(y)-sin(xy) - y sin(x)] (2u + v)

Am I correct in both of those assessments?

11. Nov 3, 2013

### haruspex

No, your ∂x/∂v is still wrong. What, by definition, is ∂u/∂v?
And I hope you mean [-(y)sin(xy) - y sin(x)]

12. Nov 3, 2013

### Alex Bard

∂u/∂v means the change in u in respect to change in v.

But isn't it ∂x/∂v? As in the change in x in respect to v? so it would be just 1? considering the u^2 is a constant it would be 0 and v would become 1, correct?

13. Nov 3, 2013

### haruspex

Yes! ∂g/∂v means the change in g in respect to change in v if u is held constant. So by definition ∂u/∂v = 0.

14. Nov 4, 2013

### Alex Bard

Okay, so in that respect, this is what the problem should look like.

Let f (x, y) = cos(xy) + y cos(x);

= [-y sin(xy) -y sin (x)](1) + [-x sin(xy) + cos(x)](-2v)

substituting u & v for their respective values of x & y:

= [ (-u-v^2)(sin((u^2+v)(u-v^2)) - (u - v^2) sin((u^2 + v)(u - v^2))] + [(-u^2+v) sin(u-v^2) + cos(u^2+v)](-2v)

substituting the given values of u & v with solved x & y values:

u = 1
v = -1
x = u^2 + v = 1^2+(-1) = 0
y = u - v^2 = 1 - (-1)^2 = 0

= [(-2) sin((0)(0)) - (0) sin((0)(0))] + [(-2) sin(0) + cos(0)] (-2(-1))

sin(0) = 0
cos(0) = 1

= cos(0)(-2v) = 1 * (2) = 2

Last edited: Nov 4, 2013
15. Nov 4, 2013

### haruspex

You've lost a factor y in the sin(x) term, and the next term is wrong too.
If we're ever going to get through this you will need to be more careful with the algebra.

16. Nov 4, 2013

### Alex Bard

Edited the above post for correction. Thank you for point out my mistake and I will be more careful going forward.

17. Nov 4, 2013

### haruspex

Sign errors. -a-b is not the same as -(a-b).
Missing x factor inside.

18. Nov 4, 2013

### Alex Bard

Thank you for your help and I believe I got the problem already solved on paper, just do not know how to mark the post as SOLVED as I can't edit my original Post

Last edited: Nov 4, 2013
19. Nov 4, 2013

### haruspex

Yes, it would be handy if there were some status flag the originator could turn on to say the thread is finished with. Another flag requesting more help would also be useful. Sometimes people post to their own ignored thread saying 'bump', but that's easily missed.