# Homework Help: Implicit differentiation

1. Jan 2, 2010

### longrob

1. The problem statement, all variables and given/known data
find $$\frac{\mathrm{d}y}{\mathrm{d}x}$$ where y is defined implicitly as a function of x

2. Relevant equations
$$x\sin(xy)=x$$

3. The attempt at a solution

$$x(\cos(xy)(x\frac{\mathrm{d}y}{\mathrm{d}x}+y))+\sin(xy)=1$$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-\sin(xy)-xy\cos(xy)}{x^2 \cos(xy)}$$

Apparently the correct answer is $$\frac{1-y\cos(xy)}{x}$$

Thanks !!

2. Jan 2, 2010

### yungman

This is a wild guess:

xsin(xy)=x=>sin(xy)=1

$$\frac{d[sin(xy)]}{dx}=0\Rightarrow cos(xy) \frac{d(xy)}{dx}=0 \Rightarrow cos(xy)[y+x\frac{dy}{dx}]=0 \Rightarrow x\frac{dy}{dx}=-y \Rightarrow \frac{dy}{dx}=-\frac{y}{x}$$

I have no confidence on this, just a though. And still have y on the right side.

Last edited: Jan 2, 2010
3. Jan 2, 2010

### longrob

I can easily see that dy/dx=-y/x when sin(xy)=0, so I'm not sure how that helps, but thanks anyway !

4. Jan 2, 2010

### HallsofIvy

What do you mean you are "not sure how that helps"? dy/dx= -y/x is the answer!

But apparently you have written something incorrectly.
$$\frac{1- ycos(xy)}{x}$$ certainly is NOT the derivative of x sin(xy)= x and it does not make sense to me that they would leave that extraneous "x" in the equation.

5. Jan 2, 2010

6. Jan 2, 2010

### longrob

OK, I've got it now. Seems I've been led astray by the "apparent" answer.

7. Jan 2, 2010

### longrob

Sorry, I am still not sure on this.....

Is it really valid to cancel x here ?

How did cos(xy) disappear ?
And why is my attempt in my initial post wrong ?

8. Jan 2, 2010

### longrob

OK, more on this. Finally I think I have it. Both methods result in -y/x. In the simple method where we cancel x first, we have the proviso that x is not equal to 0 and in the method I wrote initially, my "result" also relies on x not equal to 0 (and also cos(xy) not equal to zero). Once this assumption is made then my result simplifies to -y/x

Glad that's out of the way.

9. Jan 2, 2010

### yungman

You expand it out, move one term to the right side, cos(xy) is on both side and can be cancel just like x on your original equation. I only wrote down every other step to save me typing all the Latex!!!! Write it out, you'll see.

One thing still bug me is there is still y on the right side and the question specified y is a function of x so I expect there is no y on the right side!!!!

10. Jan 2, 2010

### Dick

There's nothing wrong with dy/dx=(-y/x) as the solution to an implicit differentiation. To really find y(x) you could solve that differential equation. Or better yet just look back at sin(xy)=1. That means xy is a constant, right? It really seems odd to have that x on both sides of the problem definition. I suspect it's a typo.

11. Jan 2, 2010

### yungman

I thought about that last night already, $$xy=\frac{2n\pi}{2}\Rightarrow y=x\frac{2n\pi}{2}$$.

So $$\frac{dy}{dx}=-\frac{2n\pi}{2}$$

It just don't look like a normal answer to me!!!! Or is it just me??!!

12. Jan 2, 2010

### diazona

$$xy = n\pi$$
becomes
$$y = \frac{n\pi}{x}$$

13. Jan 2, 2010

### yungman

I am confussing myself!!!!!

For $$sin(xy)=1 \Rightarrow xy=\frac{\pi}{2}$$

This is the only value in each cycle of 2\pi. Therefore:

$$xy=\frac{\pi}{2}+2n\pi$$ n=0,1,2.....

$$\frac{dy}{dx}=-\frac{y}{x}=-\frac{(\frac{\pi}{2}+2n\pi)}{x^{2}}$$

Am I getting this? I seem to have a mental block on this!!!!

14. Jan 2, 2010

### Dick

Actually, you want y=(2n+1/2)*pi/x. It's easy enough to check that dy/dx=-y/x alright. Still a funny problem for implicit differentiation.

15. Jan 2, 2010

I know!!!