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1) differentiate left side 2) differentiate right side 3) isolate y' 4) substitute y (if possible)
can you give an example @lgbasallote ?
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
a good example would be \[y = a^x\] classic
note: a is a constant and x is the variable you wanna differentiate
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 :-(
for example: ysinx+x^2y^2=2x
first term -> use product rule second term -> use product rule derivative of 2x would be 2
think "\(y=f(x), y'=f'(x)\) and use the product and chain rules where needed
product rule on ysin x would be... \[y(\cos x) + y' \sin x\] try product rule on second term
when i said second term i meant X^2 y^2
so..... dy/dx sinx+ycosx+2x2y=2?
not exactly....you use power rule on x^2 y^2
*facepalm* product rule...darn these p's
so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?
not exactly...second term would be x^2 (2y)(dy/dx)
now combine all the info
wait...how is y^2 becoming 2y if your doing the product rule (f'g+fg')...
second term would be x^2 multipled to derivative of y^2 right?
derivative of y^2 is 2y (dy/dx)
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?
y^2 stayed the same in 2xy^2
in the second term x^2 remains the same
im still not getting the concept (im a bit slow tonight, sorry)
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) ?"
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
derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)
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
that's pretty much it....so now combine all the info
i mean add the derivative of ysinx and the derivative of x^2y^2
so its: dy/dx sinx+ycosx+2xy^2+x^2 2y(dy/dx)=2
yep you got it...now bring to the right side all the terms that doesnt include dy/dx
sooo dy/dx sinx+x^2 2y(dy/dx)=2-ycosx-2xy^2 so far
so is it: dy/dx=2-ycosx-2xy^2/sinx+x^2 2y ?
yep...i dont suppose you can put y in terms of x so i guess that's the final answer...can you confirm @satellite73 ?
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.