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Implicit Differentiation and the Chain Rule

by Peter G.
Tags: chain, differentiation, implicit, rule
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Peter G.
#1
Jun26-13, 05:22 PM
P: 437
Hi,

I was trying to understand why the chain rule is needed to differentiate expressions implicitly.

I began by analyzing the equation used by most websites I visited:

e.g. x2+y2 = 10

After a lot of thinking, I got to a reasoning that satisfied me... Here it goes:

From my understanding, the variable y is a function of x. This function of x is being squared. This means that we can think of f(x) as part of another function (e.g. u = g(y) = y^2). Hence, y^2 is a composite function and, thus, differentiating it would require the chain rule.

However, after coming across some different type of questions I am no longer sure my train of thought is valid. For example:

6x^2+17y = 0.

I have read that to differentiate 17 y with respect to x we also have to apply the chain rule. This does not fit with my original reasoning (since, to my eyes, y cannot be thought of as a composite function in this case)

Can anyone please help me understand why we have to use the chain rule to differentiatie implicitly?

Thank you in advance!
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Mark44
#2
Jun26-13, 05:33 PM
Mentor
P: 21,214
Quote Quote by Peter G. View Post
Hi,

I was trying to understand why the chain rule is needed to differentiate expressions implicitly.

I began by analyzing the equation used by most websites I visited:

e.g. x2+y2 = 10

After a lot of thinking, I got to a reasoning that satisfied me... Here it goes:

From my understanding, the variable y is a function of x. This function of x is being squared. This means that we can think of f(x) as part of another function (e.g. u = g(y) = y^2). Hence, y^2 is a composite function and, thus, differentiating it would require the chain rule.

However, after coming across some different type of questions I am no longer sure my train of thought is valid. For example:

6x^2+17y = 0.

I have read that to differentiate 17 y with respect to x we also have to apply the chain rule. This does not fit with my original reasoning (since, to my eyes, y cannot be thought of as a composite function in this case)
No need to apply the chain rule here. You're just differentiating y with respect to x (to get dy/dx). In your previous example, you did need to use the chain rule, since d/dx(y2) = d/dy(y2) * dy/dx. Here we had a function of y, and y itself was a function of x.
Quote Quote by Peter G. View Post

Can anyone please help me understand why we have to use the chain rule to differentiatie implicitly?

Thank you in advance!
Peter G.
#3
Jun26-13, 06:38 PM
P: 437
Oh, perfect, that is great news! I must have misread the solution to the last problem. Thank you very much!

shortydeb
#4
Jun27-13, 03:20 PM
P: 29
Implicit Differentiation and the Chain Rule

Quote Quote by Peter G. View Post

I have read that to differentiate 17 y with respect to x we also have to apply the chain rule. This does not fit with my original reasoning (since, to my eyes, y cannot be thought of as a composite function in this case)
You can think of y as a composite function, if you think of x as a function of x.
shortydeb
#5
Jun27-13, 03:38 PM
P: 29
Quote Quote by Mark44 View Post
No need to apply the chain rule here. You're just differentiating y with respect to x (to get dy/dx). In your previous example, you did need to use the chain rule, since d/dx(y2) = d/dy(y2) * dy/dx. Here we had a function of y, and y itself was a function of x.
Technically you do use the chain rule, because 17y is a function of y.
pasmith
#6
Jun27-13, 03:43 PM
HW Helper
Thanks
P: 946
Quote Quote by Peter G. View Post
I have read that to differentiate 17 y with respect to x we also have to apply the chain rule.
Or by using the product rule, together with the fact that
[tex]
\frac{\mathrm{d}(17)}{\mathrm{d}x} = 0.
[/tex]
Mark44
#7
Jun27-13, 03:56 PM
Mentor
P: 21,214
Quote Quote by shortydeb View Post
Technically you do use the chain rule, because 17y is a function of y.
Just because 17y is a function of y doesn't mean that the chain rule needs to be used.
Quote Quote by pasmith View Post
Or by using the product rule, together with the fact that
[tex]
\frac{\mathrm{d}(17)}{\mathrm{d}x} = 0.
[/tex]
Or better yet, the constant multiple rule. d/dx(17y) = 17 * dy/dx.


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