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

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.21) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you give an example @lgbasallote ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0simplest 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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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 :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for example: ysinx+x^2y^2=2x

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so..... dy/dx sinx+ycosx+2x2y=2?

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

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

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

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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what 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?

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0im still not getting the concept (im a bit slow tonight, sorry)

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.2by "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) ?"

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.22xy^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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhh 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

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

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

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

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0NICEEE :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.
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