Calculus: Implicit Differentiation

[Xaven]

Ah! Why doesn't it work!? Examples are a good way to see and learn, I suppose. Can someone implicitly differentiate, oh, for example:

y^2=(x^2-9)/(x^2+9)

? I have:

2y(dy/dx)=2x/(2x)
2y(dy/dx)=1
dy/dx=1/(2y)

But my book doesn't agree with me. Wah!

[Odaisé Gaelach]

Assuming that what I've done is not complete and utter bollocks...

y^2 = (x^2 - 9) / (x^2 + 9)

y = [(x^2 - 9) / (x^2 + 9)]^1/2 - (Square root of the right hand side)

So, using the chain rule and the quotient rule:

dy/dx = [1/2] . [[(x^2 - 9) / (x^2 + 9)]^-1/2] . [36x / (x^4+18x^2+81)]

Or, see attachment.

All that you need to realise that if y^2 = something, then y = the squared root of something. After that, it all falls into place and becomes just your everyday chain rule calculus. :)

[o_O]

The problem with your original working is that you differentiated the quotient without using the quotient rule. You can't just differentiate the numerator and the denominator separately. Here's my working:

y^2 = (x^2 - 9) / (x^2 + 9)

First:
let <i>f</i> = x^2 - 9
let <i>g</i> = x^2 + 9
let <i>h</i> = y^2
Observe that with respect to x:
f' = 2x
g' = 2x
h' = 2y(dy/dx)
By the quotient rule:
h' = (gf' - fg')/g^2

Therefore:
2y(dy/dx) = [2x(x^2+9) - 2x(x^2-9)]/(x^2+9)^2

And finally, simplifying:
2y(dy/dx) = 36x / (x^2+9)^2
y(dy/dx) = 36x / 2(x^2+9)^2

dy/dx = 18x / y(x^2+9)^2

Take that with a grain of salt - I haven't done implicit differentiation in several years. :p What does your book say?

[rubah]

That's what I got, Mike!

An image (because text sucks to understand math)

<img src="http://www.snowy-day.net/stuff/implicitdiff.png"/>

[blackmage_nuke]

Wouldnt having y in you final answer make it not finished?

I got Odaisé Gaelach's answer, but there should be a +- out the front or something

and i simplified it abit further and got (18x)/[(x^2+9)^3/2(x^2-9)^1/2]

THat being said i dont know what "implicit" means so i could be doing it completely wrong

[Flying Mullet]

Having y in the answer is fine. You're trying to find the slope of the graph at any given point with respect to x and y.

I got the same answer as o_O and rubah.

[rubah]

Since it's implicit, you don't need to get rid of both variables. You're differentiating two different things at the same time, or at least that's what I thought it was. You could probably do some voodoo with the original equation, but why bother? x.x;

[o_O]

Yes Allie, that's exactly what it means. In this case, Implicit differentiation is used because separating the equation would be a pain in the ass.

Instead of applying a regular differentiation and a quotient rule you would have to apply a regular differentiation and the quotient rule inside of the chain rule. Instead of doing a relatively simple differentiation of both sides, you end up having to differentiate y on one side and ((x^2-9)/(x^2+9))^-1 on the other. Ok, it's not really that complex a formula but for pedagogical purposes it illustrates the point well.

It's also worth pointing out that taking the square root of the right hand side before differentiation means the process is no longer implicit, just regular differentiation.

[Weimar Pluto Knight VII]

No, taking the sqrt of both sides gets you y = +/- the sqrt of the stuff on the right side. So you're have to do two problems of regular differentiation.

Only substitute y in for what y equals if you used logarithmic differentiation. For example, if you start of with y = x^x, you will use log differentiation which is taking ln's of both sides and doing implicit differentiation, but at the end you're supposed to put what y equals. But not if you started with a curve you can't solve for. Like if you get e^xy = sin(x/y) + (y^87) (x^100), I'm pretty sure you can't solve for y lol.

Another way to implictly differentiate equations with two variables is something you learn in multivariable calculus. Manipulate your equation so that you get everything = 0. So in the first example in the thread, we want y^2 -((x^2 - 9)/(x^2 +9)) = 0. Let f(x,y) = y^2 -((x^2 - 9)/(x^2 +9)). Now find -Fx/Fy, which are partial derivaties where with Fx, you just pretend y is a constant and differentiate with respect to x, and do the opposite with Fy, and you get the same thing you would if you did it the calc1 method.