Question

Use implicit differentiation to find dy/dx when 4y^2 = 3x-5/3x+5

I've done it but I keep getting the wrong answer, I must be doing something wrong but I don't know where

Answer 1

\[4y ^{2}=\frac{ 3x-5 }{ 3x+5 }\]

Answer 2

what are you getting so far?

Answer 3

well if i'd type out all my working it'd be a bit tedious but my final answer i got was (-12y^2+3) / (24xy+40y)

Answer 4

The first step is to derive both sides with respect to x. What do you get when you derive 4y^2 with respect to x?

Answer 5

well first i brought the (3x+5) to the LHS and then actually expanded it, am I allowed to expand it? so I derived 12xy^2 actually not 4y^2

Answer 6

If you did that, you would get


4y^2(3x+5) = 12xy^2+20y^2


Then you would derive 12xy^2+20y^2

Answer 7

yes

Answer 8

giving you what?

Answer 9

12y^2 + 24xy dy/dx + 40y dy/dx

Answer 10

good

Answer 11

now derive the RHS 3x-5

Answer 12

3

Answer 13

so then so far it would look like this, well this is what i got.

12y^2 + 24xy dy/dx + 40y dy/dx = 3

Answer 14

the last thing to do is isolate dy/dx

Answer 15

by taking it as a common factor from 24 and 40 terms

Answer 16

dy/dx (24xy + 40y) = -12y^2 + 3

dy/dx = -12y^2 + 3 / 24xy + 40y

Answer 17

but that's not the correct answer so I'm not sure where I went wrong?

Answer 18

I think you meant to say

\[\Large \frac{dy}{dx} = \frac{-12y^2+3}{24xy+40y}\]

Answer 19

yes

Answer 20

Keep in mind that initially


\[\large 4y^{2}=\frac{ 3x-5 }{ 3x+5 }\]


so how can we use that -12y^2 to replace it with something in terms of x?

Answer 21

sorry i'm not too sure...

Answer 22

notice how -12y^2 = -3*4y^2

Answer 23

yeh

Answer 24

so you can replace 4y^2 with the RHS of


\[\large 4y^{2}=\frac{ 3x-5 }{ 3x+5 }\]

Answer 25

and simplify

Answer 26

wait u mean, u can substitute 4y^2 = 3x-5 / 3x + 5 into our final dy/dx equation as 4y^2 right? then simplify?

Answer 27

exactly

Answer 28

it's going to be a bit messy, but it can be done

Answer 29

is this step correct?

30 / (3x+5) (24xy + 40y)

Answer 30

if 24xy+40y is in the denominator, then yes it's correct

Answer 31

you can simplify further

Answer 32

yeh, I just hate going this much in depth just for simplifying but i think i'm getting there.

Answer 33

tell me what you get

Answer 34

what does 24xy + 40y factor to?

Answer 35

8y(3x+5)

Answer 36

So (3x+5)(24xy + 40y) turns into (3x+5)*8y(3x+5) = 8y(3x+5)^2

Answer 37

Giving you


\[\large \frac{ 30 }{ 8y(3x+5)^2 }\]


\[\large \frac{ 15 }{ 4y(3x+5)^2 }\]

Answer 38

yep k, i got it.

Answer 39

thanks so much jim!

Answer 40

so in the end I was doing the right steps, just didn't simplify doing further like 100 steps...

Answer 41

here's a faster/simpler way to do it (in my opinion anyway)


\[\large 4y^{2}=\frac{ 3x-5 }{ 3x+5 }\]


\[\large \frac{d}{dx}[4y^{2}]=\frac{d}{dx}\left[\frac{ 3x-5 }{ 3x+5 }\right]\]


\[\large 8y*\frac{dy}{dx}=\frac{d}{dx}\left[\frac{ 3x-5 }{ 3x+5 }\right]\]


\[\large 8y*\frac{dy}{dx}=\frac{ 30 }{ (3x+5)^2 } \ ... \ \text{See note in next post}\]


\[\large \frac{dy}{dx}=\frac{ 30 }{ 8y(3x+5)^2 }\]


\[\large \frac{dy}{dx}=\frac{ 15 }{ 4y(3x+5)^2 }\]

Answer 42

Deriving \[\Large \frac{ 3x-5 }{ 3x+5 }\]



----------------------------------------------------------------------------------


\[\Large h(x) = \frac{ f(x) }{ g(x) } = \frac{ 3x-5 }{ 3x+5 }\]


\[\Large f(x) = 3x-5\]


\[\Large g(x) = 3x+5\]



\[\Large f^{\prime}(x) = 3\]


\[\Large g^{\prime}(x) = 3\]


----------------------------------------------------------------------------------


Quotient Rule:



\[\Large h(x) = \frac{ f(x) }{ g(x) }\]



\[\Large h^{\prime}(x) = \frac{ f^{\prime}(x)g(x) - g^{\prime}(x)f(x) }{ [g(x)]^2 }\]



\[\Large h^{\prime}(x) = \frac{ 3(3x+5)-3(3x-5) }{ (3x+5)^2 }\]



\[\large h^{\prime}(x)=\frac{ 9x+15-9x+15 }{ (3x+5)^2 }\]



\[\large h^{\prime}(x)=\frac{ 30 }{ (3x+5)^2 }\]

Answer 43

yeh that is much faster! ==

Answer 44

well Jim, thanks so much for ur help, guidance, tips and tricks! hopefully u'll be able to help me more in the future when u have time lol

Answer 45

you're welcome