# Differentiation

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Is there anything in particular you're stuck on?? I don't do the IGCSE but differentiation come up at A-level so I might be able to help :/

Well, I was wondering if somebody could explain the basics? It seems a bit silly, but I think that might be where I'm going wrong.

For example, how would you go about solving this question? :)

"A body is moving in a straight line which passes through a fixed point, 0. The displacement, s metres, of the body from O at time t seconds is given by s = t^{3} + 4t^{2} – 5t."

(a) Find an expression for the velocity, v m/s, at time t seconds

(b) Find the acceleration after 2 seconds

Thanks! :D

Well im not too sure but ...

so u know when u differentiate you got to do the whole dy/dx ... well cus ur using s and t it would be ds/dt which would mean ds/dt=3t^2+8t-5 ...... so then as it is at the fixed point 0 ... u would make the equation equal 0 and work out the values of s and t .... i got t =-1 and t=5 so then you sub it into the 1st equation to work out s :/

o god im not actually sure ... what kinda maths r u doin because i dnt recognise this stuff very much :P sorry.. the other guy will probz be better help than me :p

At A-level differentiation comes up in C3 and C4 but that questions looks more like a Mechanics question!! And if it is a Mechanics question then you shouldn't need differentiation...

But if you were to differentiate s=t^3+4t^2-5t then it would go something like this:

dS/dt = 3t^2 + 8t - 5

You take each seperate bit with t in it, multiply it by the current power then take away one from the power.

For example: t^3 => bring the 3 in front so you have 3t^3, then minus one from the power so you have 3t^2.

The same with 4t^2 => bring the 2 down and multiply it with the 4 so you have 8t^2, then minus one from the power so you have 8t^1. But obviously 8t^1 is just 8t so you don't have to write the one.

Then with 5t it's actually 5t^1 so you bring the one down and multiply it by 5, so it stays as 5t^1. Then minus one from the power so you have 5t^0. But anything to the power of 0 is just one, so it stays as just 5.

Hence, dS/dt = 3t^2+8t-5

I'm sorry this doesn't quite answer your question but I will try and work on it and in the meantime, if you do need to differentiate 's' then that is how you do it!

:)

Thanks very much :)

I think IGCSE is a bit weird to be honest: my younger brother's doing the regular Edexcel GCSE Maths a couple of years early (his school is very keen on that) but he's not doing that either. Is it a special form of punishment reserved for people doing my specification? :O

Anyway, thank you very much for your help, guys :)

differentiate s to find v then diferentiate v to find a

s = t^{3} + 4t^{2} – 5t

ds/dt= v

v= 3t^2 + 4(2t)-5 (the t disappears becoz t is actually t^1 and 1-1=0 anything to the power of 0 is 1)

v=3t^2+8t-5

now a=dv/dt

a=3(2t)+8

a=6t+8

substitute t=2s into a

a=6x2+8

a=20m/s

I hope this helps. If u still don't understand just ask. I'll try and get back to u asap.