Implicit differentiation ?

Tutoring | Bookmark Us | FREE to Join! New User? Sign Up | Sign In | Help

Students Ask Questions--Tutors Answer
TutorAnswer.com

TutorsTeach.com

ask.

answer.

discover.

My Profile

Home > Calc/Pre-calc > Resolved Questions

Resolved Questions

Implicit differentiation

Jcatta:
I saw the earlier post about implicit differentiation. I was actually asking because a student was trying to figure out how to solve the problem and I became curious about it myself. My calculus skills are a little rusty. Can you explain to me the rules of differentiation you used to come up with the second equation?
Obviously, the derivative of a constant is 0 but I\'ve forgotten how to take derivatives when x and y terms are multiplied together. If the rules have a name you can just list those and I\'ll look them up in my calc book.
If not, could you explain them to me? I have gone through the entire course of calculus before but it's been 10 years since my last class, so I just need something to jog my memory.
d/dx[xy^2] + d/dx[x^2y] = d/dx[2] => 2xydy/dx + y^2(1) + 2xy + x^2dy/dx = 0

Posted 224 day ago

Report Abuse


Doctor Bob	Best Answer My Calculus is beyond rusty but there is a terrific website which addresses your question. http://en.wikibooks.org/wiki/Calculus/More_Differentiation_Rules Good luck. Posted 224 day ago ( 0 ) ( 0 ) Report Abuse

Find Interesting

Email to Friends

Bookmark

Subscribe to Answer Alert

Answers (4)

implicit differenciation is is done as foolows:
Suppose we have y^4 + x^3y^5 - x^2 = 12.
We want to find dy/dx, of course assuming y is a function of x.
d/dx(y^4 + X^3Y^5 - x^2 ) = d/dx(12)
d/dx (y^4) + d/dx(x^3y^5)-d/dx(x^2) = 0
4y^3dy/dx +(x^3).5y^4dy/dx+(3x^2y^5)-2x=0
Then
dy/dx(4y^3 + 5x^3y^4) = (2x - 3x^2y^5),
only terms involving dy/dx are on the left side. Other terms are moved to the right.
( 2x-3x^2Y^5)
So, dy/dx = -----------------------,
(4y^3 + 5x^3y^4)
This is the implicit derivative of the given function.

Posted 224 day ago

( 0 )

( 0 )

Report Abuse

MFB13

Hi, I'm a bit rusty myself where my last calculus class which was differential equations was about 32 years ago but anyway when you are differentiating implicitly for example xy, first, you always utilize the product rule for derivatives and you also use the chain rule for derivatives. That is for example to implicitly differentiate 3(y^3)(x^4) + yx

For the first term, 3 first gets factored out and remains as 3. The rest of the first term again you would use the product rule and the chain rule (you can look up these rules in any calculus text) as follows:

3(y^3)(x^4) = 3 [ (3y^2)(dy/dx)(x^4) + (y^3)(4x^3) ]

notice that inside, I wrote dy/dx because that is part of the chain rule for the y term.
Again, lets do the second term above which is yx and that is:

yx = (dy/dx)x + y (1) = x (dy/dx) + y

Again, using the chain rule, dy/dx comes from y only.
So the final answer to both terms is:

3 [ (3y^2)(dy/dx)(x^4) + (y^3)(4x^3) ] + x (dy/dx) + y

I hope this was helpful, but if not, let me know.

Posted 224 day ago

( 0 )

( 0 )

Report Abuse

matherapist

I think this is an application of the rule for differentiating products ie d/dx(f(x)*g(x))= f(x) times the derivitive of g(x) plus the second times the derivitive of the first or f(x)g'(x)+f'(x)g(x) where ' measns d/dx of the function. this can be easily remembered by taking the derivitive of hi*ho (*= multiplication) which is equal to hidho + hodhi. So for your problem

d/dx[xy^2)= x*2y*dy/dx +y^2*(d/dx of x)
d/dx(x^2*y)= 2x(dx/dy)y + x^2*(dy/dx)
d/dx of a constant =0

Add them up and get 2xy(dy/dx) + 1*y^2 ++2xy(dx/dy)+x^2(dy/dx) = 0

Rule for differentiating x^n is n*x^(n-1)*d/dx

note that your term "2xy" needs a dx/dy after it

Posted 224 day ago

( 0 )

( 0 )

Report Abuse

Doctor Bob

My Calculus is beyond rusty but there is a terrific website which addresses your question.
http://en.wikibooks.org/wiki/Calculus/More_Differentiation_Rules
Good luck.

Posted 224 day ago

( 0 )

( 0 )

Report Abuse

©2009 TutorAnswer.com, All rights reserved! • Contact us • Terms & Conditions • Privacy Policy •