How Do You Solve This Implicit Differentiation Problem?

[longrob]

Homework Statement

find \frac{\mathrm{d}y}{\mathrm{d}x} where y is defined implicitly as a function of x

Homework Equations

x\sin(xy)=x

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 !

 

[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.

 

[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 !

 

[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.

 

[longrob]

The "apparent" answer came from
http://www.analyzemath.com/calculus/Differentiation/implicit.html
See the bottom of the page

If the actual answer is dy/dx = -y/x, please can you explain. I showed my working in my initial posting. Where did I go wrong ?

 

[longrob]

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

 

[longrob]

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

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

Is it really valid to cancel x here ?

<br /> \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}<br />

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

 

[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.

 

[yungman]

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 ?

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!

 

[Dick]

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!

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.

 

[yungman]

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.

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??!

 

[diazona]

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

 

[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!

 

[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.

 

[yungman]

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.

I know!