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Further Maths GCSE Differentiation question - need detailed instructions watch

1. I've ben testing myself for my Further Maths GCSE and as usual, harder questions than I've been taught have come up frequently in past papers - I can't understand this one at all: 'Find dy/dx when y=ax^3 - 2/x where a is a constant' Can someone give detailed instructions on how to do this and maybe explain to me what the constant a is about?
2. (Original post by studentdying999)
'Find dy/dx when y=ax^3 - 2/x where a is a constant'
If the question asked you to find dy/dx when y=4x^3 - 2/x, could you do that?

(And I don't know any maths GCSE that would include a question like this - even the further maths ones probably wouldn't include the 2/x term.)
3. (Original post by studentdying999)
I've ben testing myself for my maths GCSE and as usual, harder questions than I've been taught have come up frequently in past papers - I can't understand this one at all: 'Find dy/dx when y=ax^3 - 2/x where a is a constant' Can someone give detailed instructions on how to do this and maybe explain to me what the constant a is about?
Differentiation is not part of GCSE. Do you mean IGCSE? Which past paper were you looking at?
4. Differentiation is not part of GCSE maths. But if you plan on taking maths further to A Level you will come across loads of these questions. It's really good that your challenging yourself though but don't worry about it right now.

The general rule for differentiation is that you decrease the power by 1 and times by the original power. So if y = 2x^2 , dy/dx = 4x.

I'm not too sure what it means by a as a constant. I would guess you would just leave it along when differentiating.
5. (Original post by Pangol)
If the question asked you to find dy/dx when y=4x^3 - 2/x, could you do that?

(And I don't know any maths GCSE that would include a question like this - even the further maths ones probably wouldn't include the 2/x term.)
I could do that yes - sorry I meant to say it was a further maths past paper
6. (Original post by studentdying999)
sorry I meant to say it was a further maths past paper
7. (Original post by notnek)
Ok thank you
8. (Original post by studentdying999)
I could do that yes
In that case, you'll be fine with this.

a is just a number. In the example I gave you, a=4, so when you differentiate the first part (ignore the last part for now, if you can do that it is not related to the part that you are finding a problem), you get dy/dx = (3x4)x^2 = 12x^2. In your case, you don't know what a is, but that doesn't mean that you can't follow exactly the same method. You can still bring down the power of 3 and multiply it by a. The only thing you can do is end up with a single number here (like the 12 in my example).
9. (Original post by Pangol)
In that case, you'll be fine with this.

a is just a number. In the example I gave you, a=4, so when you differentiate the first part (ignore the last part for now, if you can do that it is not related to the part that you are finding a problem), you get dy/dx = (3x4)x^2 = 12x^2. In your case, you don't know what a is, but that doesn't mean that you can't follow exactly the same method. You can still bring down the power of 3 and multiply it by a. The only thing you can do is end up with a single number here (like the 12 in my example).
So in an exam would you use 4 (or any other number) as your constant instead or would you just differentiate a and come out with a result including a in it? (eg. 3ax^2)
10. (Original post by studentdying999)
So in an exam would you use 4 (or any other number) as your constant instead or would you just differentiate a and come out with a result including a in it? (eg. 3ax^2)
The latter - I only used the example with the 4 to help you to realise that you can already do this. 3ax^2 is the differential of the first part.
11. (Original post by Pangol)
The latter - I only used the example with the 4 to help you to realise that you can already do this. 3ax^2 is the differential of the first part.
Brilliant thank you!

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