anonymous
  • anonymous
I am having trouble understanding the concept of implicit differentiation. Can anyone explain it in simple terms?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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lgbasallote
  • lgbasallote
1) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)
anonymous
  • anonymous
can you give an example @lgbasallote ?
anonymous
  • anonymous
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

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lgbasallote
  • lgbasallote
a good example would be \[y = a^x\] classic
lgbasallote
  • lgbasallote
note: a is a constant and x is the variable you wanna differentiate
anonymous
  • anonymous
i saw that same video @satellite73 . it was okay-i tied to apply the same thing to a problem my professor gave me and i totally stalled :-(
anonymous
  • anonymous
for example: ysinx+x^2y^2=2x
lgbasallote
  • lgbasallote
first term -> use product rule second term -> use product rule derivative of 2x would be 2
anonymous
  • anonymous
think "\(y=f(x), y'=f'(x)\) and use the product and chain rules where needed
lgbasallote
  • lgbasallote
product rule on ysin x would be... \[y(\cos x) + y' \sin x\] try product rule on second term
lgbasallote
  • lgbasallote
when i said second term i meant X^2 y^2
anonymous
  • anonymous
so..... dy/dx sinx+ycosx+2x2y=2?
lgbasallote
  • lgbasallote
not exactly....you use power rule on x^2 y^2
lgbasallote
  • lgbasallote
*facepalm* product rule...darn these p's
anonymous
  • anonymous
so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?
lgbasallote
  • lgbasallote
not exactly...second term would be x^2 (2y)(dy/dx)
lgbasallote
  • lgbasallote
now combine all the info
anonymous
  • anonymous
wait...how is y^2 becoming 2y if your doing the product rule (f'g+fg')...
lgbasallote
  • lgbasallote
second term would be x^2 multipled to derivative of y^2 right?
lgbasallote
  • lgbasallote
derivative of y^2 is 2y (dy/dx)
anonymous
  • anonymous
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?
lgbasallote
  • lgbasallote
y^2 stayed the same in 2xy^2
lgbasallote
  • lgbasallote
in the second term x^2 remains the same
anonymous
  • anonymous
im still not getting the concept (im a bit slow tonight, sorry)
lgbasallote
  • lgbasallote
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) ?"
lgbasallote
  • lgbasallote
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
lgbasallote
  • lgbasallote
derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)
anonymous
  • anonymous
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
lgbasallote
  • lgbasallote
that's pretty much it....so now combine all the info
lgbasallote
  • lgbasallote
i mean add the derivative of ysinx and the derivative of x^2y^2
anonymous
  • anonymous
so its: dy/dx sinx+ycosx+2xy^2+x^2 2y(dy/dx)=2
lgbasallote
  • lgbasallote
yep you got it...now bring to the right side all the terms that doesnt include dy/dx
anonymous
  • anonymous
sooo dy/dx sinx+x^2 2y(dy/dx)=2-ycosx-2xy^2 so far
anonymous
  • anonymous
so is it: dy/dx=2-ycosx-2xy^2/sinx+x^2 2y ?
lgbasallote
  • lgbasallote
yep...i dont suppose you can put y in terms of x so i guess that's the final answer...can you confirm @satellite73 ?
anonymous
  • anonymous
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.

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