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malexander
Group Title
I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms?
 2 years ago
 2 years ago
malexander Group Title
I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms?
 2 years ago
 2 years ago

This Question is Closed

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
1) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
can you give an example @lgbasallote ?
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
simplest explanation i have ever seen is in the video here http://patrickjmt.com/ i cannot link directly to the video, but it is in the calculus section and very easy to understand
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
a good example would be \[y = a^x\] classic
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
note: a is a constant and x is the variable you wanna differentiate
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
i saw that same video @satellite73 . it was okayi tied to apply the same thing to a problem my professor gave me and i totally stalled :(
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
for example: ysinx+x^2y^2=2x
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
first term > use product rule second term > use product rule derivative of 2x would be 2
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
think "\(y=f(x), y'=f'(x)\) and use the product and chain rules where needed
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
product rule on ysin x would be... \[y(\cos x) + y' \sin x\] try product rule on second term
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
when i said second term i meant X^2 y^2
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
so..... dy/dx sinx+ycosx+2x2y=2?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
not exactly....you use power rule on x^2 y^2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
*facepalm* product rule...darn these p's
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
not exactly...second term would be x^2 (2y)(dy/dx)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
now combine all the info
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
wait...how is y^2 becoming 2y if your doing the product rule (f'g+fg')...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
second term would be x^2 multipled to derivative of y^2 right?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
derivative of y^2 is 2y (dy/dx)
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
what im saying is that, if g(x) is y^2, the product rule calls for one of the g(x) not to become the g'(x)...so wouldnt one of the y^2 stay the same?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
y^2 stayed the same in 2xy^2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
in the second term x^2 remains the same
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
im still not getting the concept (im a bit slow tonight, sorry)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
by "second term" i am following your answer you said "so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?"
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
2xy^2 is correct...you held y^2 constant then took the derivative of x^2 for the second term you held x^2 constant but took the derivative of y^2 wrong
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
ohhhh okay! now i understand...so anytime you take the derivative of y...its not only its derivative but you put dy/dx at the end of it
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
that's pretty much it....so now combine all the info
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
i mean add the derivative of ysinx and the derivative of x^2y^2
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
so its: dy/dx sinx+ycosx+2xy^2+x^2 2y(dy/dx)=2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
yep you got it...now bring to the right side all the terms that doesnt include dy/dx
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
sooo dy/dx sinx+x^2 2y(dy/dx)=2ycosx2xy^2 so far
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
so is it: dy/dx=2ycosx2xy^2/sinx+x^2 2y ?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.2
yep...i dont suppose you can put y in terms of x so i guess that's the final answer...can you confirm @satellite73 ?
 2 years ago

malexander Group TitleBest ResponseYou've already chosen the best response.0
NICEEE :D I never got this concept until now. Thanks so much for helping me out with this @lgbasallote . I will not close the question until @satellite73 responds to your query.
 2 years ago
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