Question

use implicit differentiation
find y'
problem is below plss show ur solution

Answer 1

\[x ^{2}y ^{2}=x ^{2}+y ^{2}\]

Answer 2

first step is product and chain rule
\[2xy^2+2x^2yy'=2x+2yy'\]

Answer 3

second step is algebra to solve for \(y'\)

Answer 4

how did you get the 2x^2yy'?? i dont get it

Answer 5

ok the idea is this
you are thinking that \(y\) is a function of \(x\) even though you do not have it explicitly written as one
i.e. \(y=f(x)\)

Answer 6

so the expression on the left looks like
\[x^2f^2(x)\] and you take the derivative using the product rule and the chain rule

Answer 7

by the product rule you get the derivative of \(x^2f^2(x)\) is
\[2xf^2(x)+x^2\frac{d}{dx}[f^2(x)]\] that is the derivative of \(x^2\) times \(f^2(x)\) plus \(x^2\) times the derivative of \(f^2(x)\)

Answer 8

that gives
\[2xf^2(x)+x^2\times 2f(x)f'(x)\] again by the chain rule

Answer 9

now replace \(f(x)\) by \(y\) and \(f'(x)\) by \(y'\) and you get
\[2xy^2+2x^2yy'\]

Answer 10

im starting to know it

Answer 11

of course when you do such a problem you do not write out what i wrote, that is just the explanation as best as i can write here

Answer 12

a more prosaic explanation (well not really an explanation, just a "how do you do it") is found on patricks just math tutoring site under "implicit diff"

Answer 13

so \[f'(x) = y'\]

Answer 14

yes

Answer 15

ok i understand now

Answer 16

tnx

Answer 17

you are pretending \(y\) is a function of \(x\) even though you don't know explicitly what is it.
that is why it is called "implicit diff"

Answer 18

yw