# Derivative of an equation

1. Aug 15, 2007

### antinerd

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

Find dy/dx in terms of x and y

2. Relevant equations

x^2 + y^2 + ln(2) = xy

3. The attempt at a solution

I'm pretty sure I did it right, but I am wondering if I was supposed to take the derivative of everything:

2x + 2y = 1

Is that the correct derivative? If so, then I'm good to go...

2. Aug 15, 2007

### rocomath

ln(2) and xy can be derived as well

3. Aug 15, 2007

### antinerd

Didn't I already derive xy? It's 1 right?

And isn't ln(2) a constant?

4. Aug 15, 2007

### rocomath

xy = product rule ... are you doing implicit/explicit differentiation? y' if you are.

ln(2) ... i forgot, been a while since i've taken Calculus

5. Aug 15, 2007

### antinerd

It just says: "Find dy/dx"

So, OK, let me correct the dy/dx xy:

so it's y + x

Now, can the answer be written thus:

x + y = 1 ??

And would I have to write it like this:

dy/dx x+y = 1 dy/dx

6. Aug 15, 2007

### d_leet

Where is this "1" coming from? And that still isn't the correct derivative of xy, you need to use the product rule. What is the derivative of y with respect to x?

7. Aug 15, 2007

### rocomath

have you learned the product rule?

f(x)g(x) = f'(x)g(x) + g'(x)f(x)

8. Aug 15, 2007

### abelian jeff

You need to take partial derivatives, and in the process of doing this, you will get dy/dx 's in your equation. Then group like terms and solve for dy/dx. Have you learned about partial derivatives? The product rule mention above will be helpful in the process, as well.

9. Aug 15, 2007

### antinerd

Alright hold up.

10. Aug 15, 2007

### d_leet

I'd imagine this problm is showing up in the context of a calulus one class after a lesson on implicit differentiation, hence the original poster has probably never encountered partial derivatives before, and has no reason at all to find them for this problem, I don't understand why you think he needs to take partial derivatives.

11. Aug 15, 2007

### abelian jeff

Crap, you're right, I made a typo. I meant to type implicit, not partial. Apologies.

12. Aug 15, 2007

### Saladsamurai

Not to be nitpicky, but I think what rocophysics meant was
(fg)'(x)=f'(x)g(x) + g'(x)f(x)

I don't go looking to correct peoples' mistakes, however I know how difficult learning new notations can be. If we all start out on the same page, it will be easier down the road.

Casey

13. Aug 15, 2007

### antinerd

The thing that I'm not getting is how to get from

2x + 2y + ln2 = xy

to

d/dx (2x + 2y + ln2) = xy

and what to do after that, and what to do with the other side of the equation...

14. Aug 15, 2007

### antinerd

Can someone show me a small example of how to do these derivatives?

15. Aug 15, 2007

### abelian jeff

If you take the derivative with respect to x of the left hand side of the equation, you have to take it of the right hand side, too.

However, simply writing d/dx on the left side there isn't enough. You need to take the derivative with respect to x to each of the summands on the left side (and, of course, the right side of the equation too).

For instance, the derivative w.r.t. x of $$5x^3$$ would be $$15x^2(\frac{dx}{dx})=15x^2.$$
The derivative w.r.t. x of $$4y^3$$, however would be $$12y^2(\frac{dx}{dy})$$. The $$\frac{dx}{dy}$$ remains in this case.

You have learned about implicit differentiation, correct?

Last edited: Aug 15, 2007
16. Aug 15, 2007

### Saladsamurai

Let's start over. You are being asked to implicitly differentiate the function
$$x^2+y^2+ln(2)=xy$$

Now how about an easier one to start. Can you try to implicitly differentiate
$$x^2+y^2=1$$ with respect to x?

17. Aug 15, 2007

### Dick

You want d/dx(x^2 + y^2 + ln(2)) = d/dx(xy). Split this into four derivatives and I'm sure anyone here will jump on any one you get wrong.

18. Aug 15, 2007

### antinerd

Yeah, I have. But I suck at life.

So what about this for starters:

d/dx (x^2 + y^2 + ln2) = (xy) d/dx

would I then do:

2 + 2y (d/dx) + 0 = xy d/dx

???

19. Aug 15, 2007

### antinerd

$$x^2+y^2=1$$

2x + y^2 = 1

?

20. Aug 15, 2007

### antinerd

And what about finding dy/dx in terms of x AND y?

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