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# simple differentiation?? watch

1. The population x, in thousands, of red squirrels is modelled by the equation
x = a/1+kt
where t is the time in years, and a and k are constants. When ,

Show that
dx/dt = -kx^2/a

can anyone do this step by step cos i'm clearly missing something simple, I'm using the quotient rule
2. (Original post by londongirl)
The population x, in thousands, of red squirrels is modelled by the equation
x = a/1+kt
where t is the time in years, and a and k are constants. When ,
Is that 'When ,' meant to continue...?
3. (Original post by londongirl)
The population x, in thousands, of red squirrels is modelled by the equation
x = a/1+kt
where t is the time in years, and a and k are constants. When ,

Show that
dx/dt = -kx^2/a

can anyone do this step by step cos i'm clearly missing something simple, I'm using the quotient rule
x = a/(1+kt)
.. = a(1+kt)^(-1)
dx/dt = -ak(1+kt)^(-2) = -ak/(1+kt)^2

dx/dt = -ak/(1+kt)^2

but OMG! (1+kt) = a/x, so dx/dt = -ak/(a/x)^2 = -k/(a/x^2) = -kx^2/a
4. (Original post by MrTrig)
Is that 'When ,' meant to continue...?
Not necessary.
x = a/1+kt = a(1+kt)^-1
using chain rule (u = 1+kt):
dx/dt = -ak (1+kt)^-2
Now x/a = 1/(1+kt)
so dx/dt = -ak(x/a)^2 = (-kx^2)/a
QED
5. a(1+kt)^-1
Goes to
-ak(1+kt)^-2
From the first thing:
1+kt=a/x
-ak(a/x)^-2

And that rearranges to your thingy. I think xD

 Okay okay. You win this time.
But I still take the bronze. >>;;

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