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)

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- February 22nd 2010, 02:35 PMstealthmathsimplicit 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, 02:57 PMicemanfan

Using implicit differentiation and the product rule:

- February 22nd 2010, 03:00 PMArchie Meade
- February 22nd 2010, 03:22 PMstealthmaths
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, 04:13 PMSoroban
Hello, stealthmaths!

Sorry, I can't follow your work . . .

Quote:

If find an expression for

First, I would divide by 4: .

Then I would expand: .

Differentiate implicitly: .

. . . .Rearrange terms: . . . .

. . . . . . . . . . .Factor: . . . . .

. . . . . . . . Therefore: . . . . . . . . . . .

- February 23rd 2010, 05:35 PMstealthmaths
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.