Page 94 - Calculus Workbook For Dummies

78 Part III: Differentiation

What to Do with Ys: Implicit Differentiation

You use implicit differentiation when your equation isn’t in “y =” form, such as
3
3
2
siny = x + y 5 , and it’s impossible to solve for y. If you can solve for y, implicit differ-
entiation will still work, but it’s not necessary.
Implicit differentiation problems are chain rule problems in disguise. Here’s what I
mean. You know that the derivative of sinx is cosx, and that according to the chain
3 3 3 l
rule, the derivative of sinx is cos x $ _i x i . You would finish that problem by doing
_
3
the derivative of x , but I have a reason for leaving the problem unfinished.
To do implicit differentiation, all you do (sort of) is every time you see a “y” in a prob-
3 3 3 3 l
_
lem, you treat it as if it were x . Thus, because the derivative of sinx is cos x $ _i x i ,
the derivative of siny is cosy y $ l. Then, after doing the differentiation, you just
rearrange terms so that you get yl = something.
By the way, I used “y” in the explanation above, but that’s not the whole story.
dy
3
3
Consider that y = l 20 x is the same as = 20 x . It’s the variable on the top that you
dx
apply implicit differentiation to. This is typically y, but it could be any other variable.
And it’s the variable on the bottom that you treat the ordinary way. This is typically x,
but it could also be any other variable.

Q. If y + x = siny + sinx, find dy . 3. Factor out yl.
3
3
dx 2 2
l
i
A. y = l cosx - 3 x 2 y 3 _ y - cosy = cosx - 3 x
2
3 y - cosy
4. Divide.
1. Take the derivative of all four terms, cosx - 3 x 2
using the chain rule for terms contain- y = l y 3 2 - cosy
ing y and using the ordinary method
for terms containing x. That’s your answer. Note that this
derivative — unlike ordinary
3 $ 3 x = cosy y + l cosx
y y + l
2
2
$
derivatives — contains ys as well as xs.
2. Move all terms containing yl to the left
side and all other terms to the right
side.
3 $ cosy y = l cosx - 3 x 2
2
$
y y - l