# implicit differentiation

• February 22nd 2010, 01:35 PM
stealthmaths
implicit differentiation
Hi:

If 8y(x+y)=4x, find an expression for dy/dx

8y(x+y)=4x
u=8y
v=(x+y)

du/dx=8
dv/dx=1

dy/dx=8y.1dy/dx+(x+y).1=9

dy/dx=8ydy/dx+(x+y)=9

I'm not sure if my method is solid here. Would somebody please check this for me. Thanks(Happy)
• February 22nd 2010, 01:57 PM
icemanfan
$8y(x+y) = 4x$

$8xy + 8y^2 = 4x$

Using implicit differentiation and the product rule:

$8xy' + 8y + 16yy' = 4$

$8xy' + 16yy' = 4 - 8y$

$y'(8x + 16y) = 4 - 8y$

$y' = \frac{4 - 8y}{8x + 16y} = \frac{1 - 2y}{2(x + 2y)}$
• February 22nd 2010, 02:00 PM
Quote:

Originally Posted by stealthmaths
Hi:

If 8y(x+y)=4x, find an expression for dy/dx

8y(x+y)=4x uv=8y(x+y)=4x
u=8y
v=(x+y)

du/dx=8 typo....du/dy=8
dv/dx=1 dv/dx=1+dy/dx

dy/dx=8y.1dy/dx+(x+y).1=9 no!

dy/dx=8ydy/dx+(x+y)=9

I'm not sure if my method is solid here. Would somebody please check this for me. Thanks(Happy)

Hi stealthmaths,

you need to reconsider this...

$8y(x+y)=4x$

Both sides are equal.
Differentiate both sides, the derivatives are equal...

$\frac{d}{dx}[8y(x+y)]=\frac{d}{dx}[4x]$

$v\frac{du}{dx}+u\frac{dv}{dx}=4$

$u=8y,\ \frac{du}{dx}=\frac{d}{dx}[8y]=8\frac{dy}{dx}$

$v=x+y,\ \frac{dv}{dx}=\frac{d}{dx}x+\frac{d}{dx}y=1+\frac{ dy}{dx}$

$(x+y)8\frac{dy}{dx}+8y\left(1+\frac{dy}{dx}\right) =4$

$\frac{dy}{dx}\left(8(x+y)+8y\right)+8y=4$

$\frac{dy}{dx}\left(8(x+y)+8y\right)=4-8y$

$\frac{dy}{dx}=\frac{4-8y}{8x+16y}=\frac{1-2y}{2x+4y}$
• February 22nd 2010, 02:22 PM
stealthmaths
I'm really missed some fundamentals from my lectures on this. I have a video tutorial I think I will watch before I fire some questions back, if that's ok.

In the meantime: I just bought mathtype so I can put my workings in Latex for easy comprehension, and managed to create the problem in the program. But how do I get it into this window without uploading an image file?
• February 22nd 2010, 03:13 PM
Soroban
Hello, stealthmaths!

Quote:

If $8y(x+y)\:=\:4x,\$ find an expression for $\frac{dy}{dx}$

First, I would divide by 4: . $2y(x+y) \:=\:x$

Then I would expand: . $2xy + 2y^2 \:=\:x$

Differentiate implicitly: . $2x\frac{dy}{dx} + 2y + 4y\frac{dy}{dx} \:=\:1$

. . . .Rearrange terms: . . . . $2x\frac{dy}{dx} + 4y\frac{dy}{dx} \;=\;1 - 2y$

. . . . . . . . . . .Factor: . . . . . $2(x+2y)\frac{dy}{dx} \;=\;1-2y$

. . . . . . . . Therefore: . . . . . . . . . . . $\frac{dy}{dx} \;=\;\frac{1-2y}{2(x+2y)}$

• February 23rd 2010, 04:35 PM
stealthmaths
Great!
I have studied and can now understand.
Thanks to Archie Mead for the full version, as this is how we are being taught.
I have also enjoyed to learn the shorter versions thanks to Soroban and Icemanfan.