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

- anonymous

- jamiebookeater

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- lgbasallote

1) differentiate left side
2) differentiate right side
3) isolate y'
4) substitute y (if possible)

- anonymous

can you give an example @lgbasallote ?

- 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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## More answers

- lgbasallote

a good example would be \[y = a^x\] classic

- lgbasallote

note: a is a constant and x is the variable you wanna differentiate

- 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

for example: ysinx+x^2y^2=2x

- lgbasallote

first term -> use product rule
second term -> use product rule
derivative of 2x would be 2

- anonymous

think "\(y=f(x), y'=f'(x)\) and use the product and chain rules where needed

- lgbasallote

product rule on ysin x would be... \[y(\cos x) + y' \sin x\] try product rule on second term

- lgbasallote

when i said second term i meant X^2 y^2

- anonymous

so..... dy/dx sinx+ycosx+2x2y=2?

- lgbasallote

not exactly....you use power rule on x^2 y^2

- lgbasallote

*facepalm* product rule...darn these p's

- anonymous

so the product rule on x^2y^2 would be 2xy^2+x^2(dy/dx) ?

- lgbasallote

not exactly...second term would be x^2 (2y)(dy/dx)

- lgbasallote

now combine all the info

- anonymous

wait...how is y^2 becoming 2y if your doing the product rule (f'g+fg')...

- lgbasallote

second term would be x^2 multipled to derivative of y^2 right?

- lgbasallote

derivative of y^2 is 2y (dy/dx)

- 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

y^2 stayed the same in 2xy^2

- lgbasallote

in the second term x^2 remains the same

- anonymous

im still not getting the concept (im a bit slow tonight, sorry)

- 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

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

derivative of y^2 is not simply dy/dx...it's 2y(dy/dx)

- 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

that's pretty much it....so now combine all the info

- lgbasallote

i mean add the derivative of ysinx and the derivative of x^2y^2

- anonymous

so its:
dy/dx sinx+ycosx+2xy^2+x^2 2y(dy/dx)=2

- lgbasallote

yep you got it...now bring to the right side all the terms that doesnt include dy/dx

- anonymous

sooo
dy/dx sinx+x^2 2y(dy/dx)=2-ycosx-2xy^2 so far

- anonymous

so is it:
dy/dx=2-ycosx-2xy^2/sinx+x^2 2y ?

- 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

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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