Trying to do my homework...

I'm working on the chapter on Implicit Differentiation, My class notes say "used when you cant solve an equation for y and you want to take the derivative." Thats all I got. The book is no help - it was written by morons who dont understand the concept of explanation. I know it's just another way to take the derivative, but I need some sort of formula. I know there's quite a few smarties in SG land...does anyone know how to do this.

[Edited on Mar 10, 2004 by maurauder]

Que

Cockeysville Hunt Valley, MD

March 2004

MAR 10, 2004 08:50 PM

thank you for reafirming my reasons for going to art school

TygerTyger

Canada

March 2003

MAR 10, 2004 09:06 PM

Check the Math group.

nudiepix

Austin, TX

February 2004

MAR 10, 2004 09:15 PM

there wasn't a lecture on the subject?

take the derivative of your equation with respect to the variable of interest, then reduce the equation down until you can get dy/dx (or whaver differential you want).

So, if you have:
x^2+y^2 = 0
you could solve for y, then take the derivative and solve for x, or you can differentiate whatyou have:
2x +2ydy/dx = 0
dy/dx = -2x/2y
dy/dx = -x/y

i think that's right...it's been a while.

TygerTyger

Canada

March 2003

MAR 10, 2004 09:17 PM

That's about it, yup.

socrates200X

College Park, MD

February 2004

MAR 10, 2004 09:23 PM

Alright, dig this. Implicit differentiation is used when the equation you're trying to take the derivative of isn't in the form "y = whatever". Usually in this case, the equation's more like "x^2 + y^2 + 21 = 0" or something. As you can see, if you solved for y, you would get "y = (+ or -)square root of (-21 - x^2), which is hard to differentiate, esp. with that uggy (+ or -) in front.
So, we impl. diff.ate. What you really want is the derivative of y with respect to x a.k.a. f '(x) a.k.a. dy/dx. So first, you take the derivative of the whole equation with respect to x. Doing this does one of three things to the terms of the equation:

1) Suppose the term has an x in it. Then, taking the derivative with respect to x just takes the deriv. like normal. Example: d/dx(x^2) = 2x. Easy!

2) Suppose the term is a constant, like 45 or r, if you know r just stands for some number and not a variable like x. Then, taking the deriv. just zeroes it out like in all other deriv.s Example: d/dx(39) = 0, d/dx(45758494) = 0. Easier!

3) Lastly, suppose the term is a y term, like y^2. Then, you just deriv. like normal BUT you add a dy/dx after the term. Example: d/dx(y^2) = 2y dy/dx, d/dx(y ^ 5/3) = 5/3 y^2/3 dy/dx, d/dx(y) = 1 dy/dx = dy/dx. The reason you add the dy/dx is hard to explain, but to try anyway, the explanation is in braces below. Feel free to skip it, and look it over only if you're curious. Otherwise, just do as explained above: diff. normally and add dy/dx afterwards for y terms.

{When you differentiate usually, you diff. with respect to whatever variables are in the equation. Like, when you diff. the equation x^2 + 3, you do it with respect to x, meaning you're doing d/dx(x^2 + 3), which, when you follow the steps above, just gives you the normal deriv. Example: d/dx(x^2 + 3) = d/dx(x^2) + d/dx(3) = 2x + 0 = 2x. But why don't you add an dx/dx or a d3/dx after the terms like you do with y terms? You do! But dx/dx is just 1, you know, because anything divided by itself is 1. And d3/dx is just 0, because any constant deriv. is just 0. So, when you're doing "normal" differentiation, you're actually doing implicit diff. but just ignoring the dx/dx's since they're just 1, and 1 times anything is just that anything. But, with dy/dx, we can't say that's 1 since dy isn't equal to dx. So, we leave it as dy/dx.}

When you normally diff., you're trying to find dy/dx, so to find it in this case, you just solve for it after implicitly diff.ing. Example:

d/dx(x^2 + y^2 + 21 = 0) = d/dx(x^2) + d/dx(y^2) + d/dx(21) = d/dx(0).
By (1), this equals 2x + d/dx(y^2) + d/dx(21) = d/dx(0).
By (2), this equals 2x + d/dx(y^2) + 0 = 0, which equals 2x + d/dx(y^2) = 0.
By (3), this equals 2x + 2y dy/dx = 0.
Now, we solve for dy/dx to get the derivative we'e been praying for, which turns out to be dy/dx = -2x/2y = -x/y. Now, given any x and y, you can get the deriv. at that point easily. Example: At (2,3), dy/dx = -2/3.
That's implicit differentiation in a nutshell. Pretty much it's just regular diff.ing with some more work. If you have more questions about anything calc-related, feel free to mail me at necroq@hotmail.com. Make sure you head your subject with "SG: Calculus" or something, so I don't delete it like the ugly spam it just may be. Good luck!