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# C4 Differentiation Edexcel Watch

There are two k's one is K dash and then other is k.

k is defined by reference to k dash and the square root of A
they have k' as a positive constant.

they then have -k'/sqrtA times sqrt(h)

so they say that k'/sqrtA is a positive constant k, leaving -k sqrt(h)

i.e. k and k' are two different constants.
They have mentioned that

You are confusing with , both are different constants.

There are two k's one is K dash and then other is k.

k is defined by reference to k dash and the square root of A

5. (Original post by steve2005)
There are two k's one is K dash and then other is k.

k is defined by reference to k dash and the square root of A
(Original post by just george)
they have k' as a positive constant.

they then have -k'/sqrtA times sqrt(h)

so they say that k'/sqrtA is a positive constant k, leaving -k sqrt(h)

i.e. k and k' are two different constants.
(Original post by raheem94)
They have mentioned that

You are confusing with , both are different constants.

Thanks but why is k' equivalent to (-k' . A^-1/2)

Sorry i understand now thanks
6. (Original post by steve2005)
There are two k's one is K dash and then other is k.

k is defined by reference to k dash and the square root of A

thanks
7. (Original post by steve2005)
There are two k's one is K dash and then other is k.

k is defined by reference to k dash and the square root of A

Probably a typo from you.
8. (Original post by raheem94)

Probably a typo from you.
No typo.

There is nothing wrong with what I have typed.
9. (Original post by steve2005)
There are two k's one is K dash and then other is k.

k is defined by reference to k dash and the square root of A

(Original post by steve2005)
No typo.

There is nothing wrong with what I have typed.
-k

I think thats what raheem94 was trying to point out
10. (Original post by just george)
-k

I think thats what raheem94 was trying to point out
I know what he was trying to say. But what I wrote is correct because k is defined, in this case, to be negative.

Updated: March 17, 2012
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