Question

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

Answer 1

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

Answer 2

can you give an example @lgbasallote ?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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 :-(

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

now combine all the info

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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?

Answer 22

y^2 stayed the same in 2xy^2

Answer 23

in the second term x^2 remains the same

Answer 24

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

Answer 25

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) ?"

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

so its:

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

Answer 32

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

Answer 33

sooo

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

Answer 34

so is it:

dy/dx=2-ycosx-2xy^2/sinx+x^2 2y ?

Answer 35

yep...i dont suppose you can put y in terms of x so i guess that's the final answer...can you confirm @satellite73 ?

Answer 36

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.