How Do You Solve This Implicit Differentiation Problem?

In summary, implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly expressed in terms of its independent variable. It involves treating the dependent variable as an implicit function of the independent variable and using the chain rule to find its derivative. This technique is useful in solving equations and finding the slope of curves that cannot be easily differentiated using traditional methods. It is commonly used in calculus and has applications in various fields such as physics, economics, and engineering.
  • #1
longrob
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Homework Statement


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


Homework Equations


[tex]x\sin(xy)=x[/tex]


The Attempt at a Solution



[tex]x(\cos(xy)(x\frac{\mathrm{d}y}{\mathrm{d}x}+y))+\sin(xy)=1[/tex]

[tex]\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1-\sin(xy)-xy\cos(xy)}{x^2 \cos(xy)}[/tex]


Apparently the correct answer is [tex]\frac{1-y\cos(xy)}{x}[/tex]

Thanks !
 
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  • #2
This is a wild guess:

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

[tex]\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}[/tex]

I have no confidence on this, just a though. And still have y on the right side.
 
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  • #3
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
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.
[tex]\frac{1- ycos(xy)}{x}[/tex] 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.
 
  • #6
OK, I've got it now. Seems I've been led astray by the "apparent" answer.
 
  • #7
Sorry, I am still not sure on this...

xsin(xy)=x=>sin(xy)=1
Is it really valid to cancel x here ?

[tex]
\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}
[/tex]
How did cos(xy) disappear ?
And why is my attempt in my initial post wrong ?
 
  • #8
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
longrob said:
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!
 
  • #10
yungman said:
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.
 
  • #11
Dick said:
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, [tex]xy=\frac{2n\pi}{2}\Rightarrow y=x\frac{2n\pi}{2}[/tex].

So [tex]\frac{dy}{dx}=-\frac{2n\pi}{2}[/tex]

It just don't look like a normal answer to me! Or is it just me??!
 
  • #12
[tex]xy = n\pi[/tex]
becomes
[tex]y = \frac{n\pi}{x}[/tex]
 
  • #13
I am confussing myself!

For [tex]sin(xy)=1 \Rightarrow xy=\frac{\pi}{2}[/tex]

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

[tex]xy=\frac{\pi}{2}+2n\pi[/tex] n=0,1,2...

[tex]\frac{dy}{dx}=-\frac{y}{x}=-\frac{(\frac{\pi}{2}+2n\pi)}{x^{2}}[/tex]

Am I getting this? I seem to have a mental block on this!
 
  • #14
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
Dick said:
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!
 

FAQ: How Do You Solve This Implicit Differentiation Problem?

What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function when it is not possible to express the dependent variable explicitly in terms of the independent variable. This is often the case when the function is defined implicitly, meaning the relationship between the variables is not explicitly stated.

How is implicit differentiation different from standard differentiation?

Implicit differentiation uses the chain rule and product rule to differentiate functions with multiple variables, whereas standard differentiation only deals with functions that are explicitly defined. Implicit differentiation is also used when the independent variable cannot be easily isolated in the equation, while standard differentiation typically involves isolating the independent variable before taking the derivative.

When should implicit differentiation be used?

Implicit differentiation should be used when the function is defined implicitly, meaning the relationship between the variables is not explicitly stated. It is also used when the independent variable cannot be easily isolated in the equation or when the function has multiple variables.

What are some real-world applications of implicit differentiation?

Implicit differentiation is commonly used in physics and engineering to analyze physical systems and to solve problems involving rates of change. It is also used in economics and finance to model relationships between variables and to calculate marginal changes.

What are some common mistakes to avoid when using implicit differentiation?

Some common mistakes to avoid when using implicit differentiation include forgetting to use the chain rule or product rule, not properly isolating the dependent variable before taking the derivative, and incorrectly applying the power rule. It is also important to check for extraneous solutions when solving for the derivative using implicit differentiation.

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