Question

Find dy/dx by implicit differentiation if \[(x^2+y^2)^2-(\pi)x^2y=0\]

Answer 1

@ganeshie8

Answer 2

you can use the chain rule

Answer 3

$$
\large 2(x^2 + y^2) (2x + 2y ~ y ~' )- \pi(2x y + x^2~ y ' ) = 0

$$

Answer 4

okay

Answer 5

you can also write it as
$$
\large 2(x^2 + y^2) (2x + 2y ~ \dfrac{dy}{dx} )- \pi(2x y + x^2\frac{dy}{dx} ) = 0

$$

Answer 6

what do i do after? @perl

Answer 7

@ganeshie8

Answer 8

@rational

Answer 9

solve \(y'\)

Answer 10

how do we solve y'?

Answer 11

\[\large 2(x^2 + y^2) (2x + 2y ~ \color{red}{y '} )- \pi(2x y + x^2~ \color{red}{y '} ) = 0\]

Answer 12

you're done with calculus part... isolating \(\color{red}{y'}\) is just algebra

Answer 13

how not sure how...

Answer 14

do i expand it?

Answer 15

yes

Answer 16

\[2(x^2 + y^2) (2x + 2y ~ \color{red}{y '} )- \pi(2x y + x^2~ \color{red}{y '} ) = 0\]

\[2(x^2 + y^2) (2x) + 2(x^2 + y^2) (2y ~ \color{red}{y '} )- \pi(2x y) -\pi( x^2~ \color{red}{y '} ) = 0\]

Answer 17

how did you that?

Answer 18

distributed

Answer 19

\[(a+b)(c+d) = (a+b)c + (a+b)d\]

Answer 20

how did you get c and d

Answer 21

that was just an example

Answer 22

(2x) and (2y y')?

Answer 23

You have this equation to start with : \[(x^2+y^2)^2-(\pi)x^2y=0\]

implicitly differentiate with respect to x, you get :
\[2(x^2 + y^2) (2x + 2y ~ \color{red}{y '} )- \pi(2x y + x^2~ \color{red}{y '} ) = 0\]


you get this far ? :)

Answer 24

yes

Answer 25

good, next expand everything. what do you get ?

Answer 26

https://www.khanacademy.org/math/algebra-basics/quadratics-polynomials-topic/multiplying-binomials-core-algebra/v/multiplying-binomials

Answer 27

\[2(2x^3+2x^2y y'+2xy'+y^2y')-(\pi)2xy+(\pi)x^2y'=0\]
i got this, it looks weird.

Answer 28

oh i see, what you were doing. You first distribute the entire first term to the another.

Answer 29

@ganeshie8

Answer 30

yes

Answer 31

\[2(x^2 + y^2) (2x + 2y ~ \color{red}{y '} )- \pi(2x y + x^2~ \color{red}{y '} ) = 0\]

\[2(x^2 + y^2) (2x) + 2(x^2 + y^2) (2y ~ \color{red}{y '} )- \pi(2x y) -\pi( x^2~ \color{red}{y '} ) = 0\]

collect \(\color{red}{y'}\) terms

\[2(x^2 + y^2) (2x) - \pi(2x y) + \color{red}{y '} [2(x^2 + y^2) (2y ~ )-\pi( x^2)] = 0\]

\[ \color{red}{y '} [2(x^2 + y^2) (2y ~ )-\pi( x^2)] = \pi(2x y) -2(x^2 + y^2) (2x)\]

\[ \color{red}{y '} = \dfrac{ \pi(2x y) -2(x^2 + y^2) (2x)}{ 2(x^2 + y^2) (2y ~ )-\pi( x^2)}\]

Answer 32

simplify if you want to

Answer 33

So what you did to collect y' terms, you common factor? @ganeshie8

Answer 34

i'm confused. How to simplify?
The answer from the text says:
\[\frac{ 2x((\pi)y-2x^2-2y^2) }{ 4x^2y+4y^3-(\pi)x^2 }\]

Answer 35

@ganeshie8

Answer 36

\[\color{red}{y '} = \dfrac{ \pi(2x y) -2(x^2 + y^2) (2x)}{ 2(x^2 + y^2) (2y ~ )-\pi( x^2)}\]

expand terms in the denominator

Answer 37

i got the denominator, i can't get the numerator

Answer 38

\[\color{red}{y '} = \dfrac{ \pi(\color{purple}{2x} y) -2(x^2 + y^2) (\color{purple}{2x})}{ 2(x^2 + y^2) (2y ~ )-\pi( x^2)}\]

factor out \(\color{purple}{2x}\) from numerator

Answer 39

you can do that?

Answer 40

why not

Answer 41

it looks strange to me because (2xy) are together?

Answer 42

\[\large a\heartsuit \spadesuit + b\heartsuit = \heartsuit(a\spadesuit + b)\]

Answer 43

oh, i see your point :)

Answer 44

thank you too much for helping me and answering my silly questions!

Answer 45

yw:)

Answer 46

there are no silly questions, only silly answers :P