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# Differentiation AS maths

1. (Original post by aliiceconnor)
It's C1
LOL and I got an A at C1. Keep up the good work. I would try be of more use but I see Sean is doing a great job as always and beat me to it.
2. (Original post by aliiceconnor)
So f'(-2)=-4

-4=-4a+b-2a
-8=-8a+2b-4a

4a-2b+13=0
-12a+2b+8=0

4a+13-12a+8=0
-8a+21=0
f'(x) = 2ax + b, not 2ax + b - 2a. can you see why?
3. (Original post by SeanFM)
f'(x) = 2ax + b, not 2ax + b - 2a. can you see why?
Not really,
Would it be
4a+13-8a+8=0
So -4a+21=0?
4. (Original post by Epistemolog y)
LOL and I got an A at C1. Keep up the good work. I would try be of more use but I see Sean is doing a great job as always and beat me to it.
Thanks haha
5. (Original post by aliiceconnor)
Not really,
Would it be
4a+13-8a+8=0
So -4a+21=0?
I see a 13 in there so I see you're referring to the first equation you have set up which is correct.. but you are not taking in what I am saying about the a^2 that you put in there and how we are yet to set up the second equation before we can solve them simultaneously.

We have that y = ax^2 + bx + a^2. When we differentiate this to find f'(x) you can differentiate individually ax^2, bx and a^2 * x^0 to give 2ax, b and 0, so f'(x) = 2ax + b + 0. Hence f'(-2) = 2 * a * -2 + b, and so f'(-2) = b - 4a and you know what f'(-2) is. Let me know if any of that is unclear
6. (Original post by SeanFM)
I see a 13 in there so I see you're referring to the first equation you have set up which is correct.. but you are not taking in what I am saying about the a^2 that you put in there and how we are yet to set up the second equation before we can solve them simultaneously.

We have that y = ax^2 + bx + a^2. When we differentiate this to find f'(x) you can differentiate individually ax^2, bx and a^2 * x^0 to give 2ax, b and 0, so f'(x) = 2ax + b + 0. Hence f'(-2) = 2 * a * -2 + b, and so f'(-2) = b - 4a and you know what f'(-2) is. Let me know if any of that is unclear
That's what I did, I sent the next step when I tried to solve it
7. (Original post by aliiceconnor)
That's what I did, I sent the next step when I tried to solve it
If ever you set up two simultaneous equations and then solve then and then the solutions satisfy both equations (i.e put them back in to both equation 1 and 2), then assuming your equations are correct your answer is correct.
8. (Original post by SeanFM)
If ever you set up two simultaneous equations and then solve then and then the solutions satisfy both equations (i.e put them back in to both equation 1 and 2), then assuming your equations are correct your answer is correct.
Okay, thank you so much for all your help!

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