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Differentiation ,Confused:| C3

ImageUploadedByStudent Room1388446926.362951.jpg

Can somebody help me on the question above.
I got 22/81.The correct answer is 58/81 and you get that by using the quotient rule to differentiate the 2nd term.What I don't understand is why you use the quotient rule and can't turn 4x/2x+1 into 4x(2x+1)^-1 (this that impossible) ?? :s-smilie:


Posted from TSR Mobile
Reply 1
Original post by livealittle
can't turn 4x/2x+1 into 4x(2x+1)^-1



You can, that is what the quotient rule does
Reply 2
You can do that.
Reply 3
Original post by TenOfThem
You can, that is what the quotient rule does


If I do that I get the wrong answer:/
I don't get why you only get the right answer when you differentiate it using the quotient rule:frown:


Posted from TSR Mobile
Original post by livealittle
If I do that I get the wrong answer:/
I don't get why you only get the right answer when you differentiate it using the quotient rule:frown:


Posted from TSR Mobile

You've either done it incorrectly or just need to manipulate the answer.
Sometimes the answer is correct but in the wrong form.
Reply 5
Original post by livealittle
If I do that I get the wrong answer:/
I don't get why you only get the right answer when you differentiate it using the quotient rule:frown:


Posted from TSR Mobile


You need to apply the product rule and the chain rule if you bring it to the top.
Reply 6
Original post by livealittle
If I do that I get the wrong answer:/
I don't get why you only get the right answer when you differentiate it using the quotient rule:frown:


Posted from TSR Mobile


You should get the same answer whichever method you use!

You can also save yourself a bit of work by noting that

4x2x+1=4x+222x+1=222x+1\displaystyle \dfrac{4x}{2x + 1} = \dfrac{4x + 2 - 2}{2x + 1} = 2 - \dfrac{2}{2x + 1}

which is a bit easier to differentiate :smile:
Reply 7
Original post by livealittle
If I do that I get the wrong answer:/
I don't get why you only get the right answer when you differentiate it using the quotient rule:frown:


I get the correct answer either way

Do you want to post your working and we can see where you are going wrong
Reply 8
Original post by TenOfThem
I get the correct answer either way

Do you want to post your working and we can see where you are going wrong


ImageUploadedByStudent Room1388505936.829384.jpg

Sorry if I'm being a pain:frown:
What I did originally is in black.maybe I differentiated it wrongly :/ idk
The bit in blue is what I did after finding out you had to use the quotient rule.
Thanks :smile:


Posted from TSR Mobile
Reply 9
Original post by livealittle
ImageUploadedByStudent Room1388505936.829384.jpg

Sorry if I'm being a pain:frown:
What I did originally is in black.maybe I differentiated it wrongly :/ idk
The bit in blue is what I did after finding out you had to use the quotient rule.
Thanks :smile:


Posted from TSR Mobile


You need the product rule with your approach

ddx4x(2x+1)1=4(2x+1)1+(4x)(1)(2)(2x+1)2\dfrac{d}{dx} 4x(2x+1)^{-1} = 4(2x+1)^{-1} + (4x)(-1)(2)(2x+1)^{-2}

Not sure what you did - I think that you just ignored the first part
(edited 10 years ago)
Reply 10
Original post by TenOfThem
You need the product rule with your approach

ddx4x(2x+1)1=4(2x+1)1+(4x)(1)(2)(2x+1)2\dfrac{d}{dx} 4x(2x+1)^{-1} = 4(2x+1)^{-1} + (4x)(-1)(2)(2x+1)^{-2}

Not sure what you did - I think that you just ignored the first part


Ohhh yeahh its a function of a function .I can be stupid sometimes.Thankyou:smile: I get it now


Posted from TSR Mobile
Original post by livealittle
Ohhh yeahh its a function of a function ... I get it now


I do not think that you do - you did the "chain rule" part originally - it was the product rule that you did not do
Reply 12
Original post by TenOfThem
I do not think that you do - you did the "chain rule" part originally - it was the product rule that you did not do


Haha Dw ,I can do it but I didn't recognise it :|



Posted from TSR Mobile
Original post by livealittle
Haha Dw ,I can do it but I didn't recognise it :|



Ok, maybe it is just the terminology that you do not understand

But, as long as you are sure that you now understand nth the product and chain rules and when each should be applied, great

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