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Core Maths 3: Differentiation using the quotient rule

Hi there

Please could someone tell me if the below is correct;



Many thanks
Jackie
I haven't checked the algebra but how did you go from the fraction at the bottom left to the next stage ? The question doesn't give us x.
Reply 2
Avatar for raheem94
raheem94
13
Original post by jackie11
Hi there

Please could someone tell me if the below is correct;

Many thanks
Jackie


According to the question you should leave your answer as dydx=x2+2x−1x2+2x+1 \displaystyle \frac{dy}{dx}=\frac{x^2+2x-1}{x^2+2x+1}

Why did you wrote 1+1−11+1+1 ? \displaystyle \frac{1+1-1}{1+1+1} \ ?
(edited 12 years ago)
Original post by raheem94
According to the question you should leave your answer as dydx=x2+2x−1x2+2x+1 \displaystyle \frac{dy}{dx}=\frac{x^2+2x-1}{x^2+2x+1}

Why did you sub in x=1?


He didn't sub in x=1 otherwise the top would be 1 + 2 - 1 = 2

He could make into a proper fraction I guess.
Reply 4
Avatar for raheem94
raheem94
13
Original post by member910132
He didn't sub in x=1 otherwise the top would be 1 + 2 - 1 = 2

He could make into a proper fraction I guess.


Thanks for correcting me, do you understand what he was trying to do?

This can easily be made into a proper fraction, but i don't think he was trying to do this.
Reply 5
Avatar for jackie11
jackie11
OP
4
Original post by raheem94
According to the question you should leave your answer as dydx=x2+2x−1x2+2x+1 \displaystyle \frac{dy}{dx}=\frac{x^2+2x-1}{x^2+2x+1}

Why did you sub in x=1?


ok thankyou, no I just assumed that when you have the same thing on the top and bottom, i.e.

10/10, it becomes 1/1

So that's what I did for x^2/x^2 and 2x/2x
Reply 6
Avatar for TenOfThem
TenOfThem
18
Original post by jackie11
Hi there

Please could someone tell me if the below is correct;



Many thanks
Jackie


Jackie, Jackie, Jackie

You CANNOT do that

You CANNOT cancel the x^2 and then cancel the 2x's ... NO NO NO NO
Reply 7
Avatar for raheem94
raheem94
13
Original post by jackie11
ok thankyou, no I just assumed that when you have the same thing on the top and bottom, i.e.

10/10, it becomes 1/1

So that's what I did for x^2/x^2 and 2x/2x


10/10 becomes 1/1 but here the top has -1 and the bottom has +1.

If you want to simplify it, then you can do it in the following way,
x2+2x−1x2+2x+1=x2+2x+1−1−1x2+2x+1=x2+2x+1x2+2x+1−2x2+2x+1=1−2x2+2x+1 \displaystyle \frac{x^2+2x-1}{x^2+2x+1}=\frac{x^2+2x+1-1-1}{x^2+2x+1}=\frac{x^2+2x+1}{x^2+2x+1}-\frac2{x^2+2x+1}= \\ 1-\frac2{x^2+2x+1}
(edited 12 years ago)
Original post by jackie11
ok thankyou, no I just assumed that when you have the same thing on the top and bottom, i.e.

10/10, it becomes 1/1

So that's what I did for x^2/x^2 and 2x/2x


No you can't do that. say if we have x+3x+5\dfrac{x+3}{x+5}, that ISN'T equal to 1+31+5\dfrac{1+3}{1+5}

You can only simplify a fraction if the top and bottom are multiplied by the same factor, not 'added on' by the same amount.

E.g. the above isn't allowed, but this is:

x(x+3)x(x+5)=1(x+3)1(x+5)\dfrac{x(x+3)}{x(x+5)} = \dfrac{1(x+3)}{1(x+5)}

EDIT: I'm sure you probably do know this, and are just getting confused but hopefully the examples helped :smile:
(edited 12 years ago)
Reply 9
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jackie11
OP
4
aww thank you all so much :smile:

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