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I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms?
 one year ago
 one year ago
I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms?
 one year ago
 one year ago

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lgbasalloteBest ResponseYou've already chosen the best response.2
1) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)
 one year ago

malexanderBest ResponseYou've already chosen the best response.0
can you give an example @lgbasallote ?
 one year ago

satellite73Best 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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
a good example would be \[y = a^x\] classic
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
note: a is a constant and x is the variable you wanna differentiate
 one year ago

malexanderBest 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 :(
 one year ago

malexanderBest ResponseYou've already chosen the best response.0
for example: ysinx+x^2y^2=2x
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
first term > use product rule second term > use product rule derivative of 2x would be 2
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
think "\(y=f(x), y'=f'(x)\) and use the product and chain rules where needed
 one year ago

lgbasalloteBest 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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
when i said second term i meant X^2 y^2
 one year ago

malexanderBest ResponseYou've already chosen the best response.0
so..... dy/dx sinx+ycosx+2x2y=2?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
not exactly....you use power rule on x^2 y^2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
*facepalm* product rule...darn these p's
 one year ago

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

lgbasalloteBest ResponseYou've already chosen the best response.2
not exactly...second term would be x^2 (2y)(dy/dx)
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
now combine all the info
 one year ago

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

lgbasalloteBest ResponseYou've already chosen the best response.2
second term would be x^2 multipled to derivative of y^2 right?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
derivative of y^2 is 2y (dy/dx)
 one year ago

malexanderBest 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?
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
y^2 stayed the same in 2xy^2
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
in the second term x^2 remains the same
 one year ago

malexanderBest ResponseYou've already chosen the best response.0
im still not getting the concept (im a bit slow tonight, sorry)
 one year ago

lgbasalloteBest 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) ?"
 one year ago

lgbasalloteBest 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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)
 one year ago

malexanderBest 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
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
that's pretty much it....so now combine all the info
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.2
i mean add the derivative of ysinx and the derivative of x^2y^2
 one year ago

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

lgbasalloteBest 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
 one year ago

malexanderBest ResponseYou've already chosen the best response.0
sooo dy/dx sinx+x^2 2y(dy/dx)=2ycosx2xy^2 so far
 one year ago

malexanderBest ResponseYou've already chosen the best response.0
so is it: dy/dx=2ycosx2xy^2/sinx+x^2 2y ?
 one year ago

lgbasalloteBest 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 ?
 one year ago

malexanderBest 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.
 one year ago
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