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# P3-Differentiation watch

1. Write down grad ar point (1,1) at curve
y=2/(3-x^2)

Maybe I am really tired and being stupid but...if i use quotient rule so..

u=2 => du/dx=???? ...do i just put zero?

It also says write down so it seems to be implying its v easy...i am being daft and missing something..ahhh help my tired brain!!
2. -2
3. (Original post by shiny)
-2
Write down grad ar point (1,1) at curve
y=2/(3-x^2)

Maybe I am really tired and being stupid but...if i use quotient rule so..

u=2 => du/dx=???? ...do i just put zero?

It also says write down so it seems to be implying its v easy...i am being daft and missing something..ahhh help my tired brain!!
You could take the constant 2 outside the diiferentiation so you have y = 2(1/3-x²), and differentiate the bracket...
5. (Original post by ZJuwelH)
You could take the constant 2 outside the diiferentiation so you have y = 2(1/3-x²), and differentiate the bracket...
lol, ZJuwelH, you're a star, and I have to stop doing maths after 10pm
lol, ZJuwelH, you're a star, and I have to stop doing maths after 10pm
I could do maths all day (please no external force ever hold me to that)
Oops ... missed the square!

Yes its 1

dy/dx = d/dx { 2 / (3 - x^2) } = 4x / (3 - x^2)^2

At x = 1; dy/dx = 1!
Write down grad ar point (1,1) at curve
y=2/(3-x^2)

Maybe I am really tired and being stupid but...if i use quotient rule so..

u=2 => du/dx=???? ...do i just put zero?

It also says write down so it seems to be implying its v easy...i am being daft and missing something..ahhh help my tired brain!!
t = (3-x^2)
dt/dx = -2x

y=2/t = 2t^-1
dy/dt = -2t^-2 = -2/t^2

dt/dx x dy/dt = dy/dx
= 4x/t^2
=4x/(3-x^2)^2
x = 1

4/(3-1)^2 = 1

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