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# Pure maths help! watch

1. Please can you help me with this question. I can't seem to find where I am going wrong.

f(x)=[(x+1)(x+9)] / [x]

Show that f'(x)= [(x+3)(x-3)] / [x2]
2. I think the problem you are having is there isn't a question
3. (Original post by wiseguy99)
I think the problem you are having is there isn't a question
4. (Original post by Alexia_17)
Please can you help me with this question. I can't seem to find where I am going wrong.

f(x)=[(x+1)(x+9)] / [x]

Show that f'(x)= [(x+3)(x-3)] / [x2]
What have you tried?

I'd start by multiplying-out the top and dividing through, then differentiate.

5. (Original post by RogerOxon)
What have you tried?

I'd start with the chain rule: f(x)=a(x)b(x), f'(x)=a'(x)b(x)+a(x)b'(x), or you could multiply out the top and divide through.

So far all I have tried is to expand the bracket to get
x2+10x+9 all over x

But I dont how that gets and x2 on the bottom. I don't think I know what you mean by the 'Chain rule'. We have only just started differentiation friday of last week.
6. So you divide all the terms by x, giving x+10+9x^-1

Then, you times this by x^2/x^2. (This is equal to 1, but times it anyway and you'll see.)

Then factorise the numerator to get (x+3)(x-3) as the numerator and x^2 as the denominator
7. (Original post by Alexia_17)
So far all I have tried is to expand the bracket to get
x2+10x+9 all over x

But I dont how that gets and x2 on the bottom. I don't think I know what you mean by the 'Chain rule'. We have only just started differentiation friday of last week.
try to simplify x2+10x+9 all over x and then differentiate term by term ....what do you get ?
8. (Original post by Chrisabraham)
So you divide all the terms by x, giving x+10+9x^-1

Then, you times this by x^2/x^2. (This is equal to 1, but times it anyway and you'll see.)

Then factorise the numerator to get (x+3)(x-3) as the numerator and x^2 as the denominator
When you multiply it does it equal, (x^3 +10x^2 + 9^-2)/x^2

also why do you multiply it by x^2/x^2
9. (Original post by TheManor)
try to simplify x2+10x+9 all over x and then differentiate term by term ....what do you get ?
1-9x^-2 ??
10. (Original post by Alexia_17)
When you multiply it does it equal, (x^3 +10x^2 + 9^-2)/x^2

also why do you multiply it by x^2/x^2
there are a few ways to do this. You started off the right way and got x^2 + 10x + 9 all over x. Now you need to do the division i.e. divide each term in your numerator by x. Then differentiate term by term ..... I wouldn't put it over x^2 just yet ...... you can do that after differentiating ....
11. (Original post by Alexia_17)
1-9x^-2 ??
this is correct .....now combine this using fractions .... clue : 1 is x^2/x^2 ....
12. (Original post by TheManor)
there are a few ways to do this. You started off the right way and got x^2 + 10x + 9 all over x. Now you need to do the division i.e. divide each term in your numerator by x. Then differentiate term by term ..... I wouldn't put it over x^2 just yet ...... you can do that after differentiating ....
So after the division I will get, x+10+9x^-1, right?

Then after differentiation you'd then get, 1-9x^-2?
13. (Original post by Alexia_17)
So after the division I will get, x+10+9x^-1, right?

Then after differentiation you'd then get, 1-9x^-2?
yes correct. Now you have to do some algebraic re-arrangement with this to get the answer. Clue : 9x^-2 means 9/x^2 and 1 is actually x^2/x^2 ......
14. Thank you all so much. I just have one more question, not differentiation, that I am stuck on and I was wondering if any of you could help?

A boy is tobogganing down a smooth slope inclined at 20 degrees to the horizontal. Find his acceleration, assuming resistances can be neglected
15. (Original post by Alexia_17)
Thank you all so much. I just have one more question, not differentiation, that I am stuck on and I was wondering if any of you could help?

A boy is tobogganing down a smooth slope inclined at 20 degrees to the horizontal. Find his acceleration, assuming resistances can be neglected
step 1 : draw a diagram and show the forces ..... the word "smooth" tells you something ....
16. (Original post by TheManor)
step 1 : draw a diagram and show the forces ..... the word "smooth" tells you something ....
step 2 : consider the direction the boy is accelerating in and write the "equation of motion" for this i.e. use Newton's 2nd law or "F=ma" ....
17. (Original post by TheManor)
step 1 : draw a diagram and show the forces ..... the word "smooth" tells you something ....
But what do you do about the mass of the boy?
smoothmeans there no friction...
18. (Original post by Alexia_17)
But what do you do about the mass of the boy?
smoothmeans there no friction...
yes - there is no friction. Don't worry about the mass just now.

The boy is sliding down the slope. What is the size of the force that is acting on him down the slope (this is where your force diagram comes in handy) .....
19. (Original post by TheManor)
yes - there is no friction. Don't worry about the mass just now.

The boy is sliding down the slope. What is the size of the force that is acting on him down the slope (this is where your force diagram comes in handy) .....
just call the mass "m"
20. (Original post by TheManor)
yes - there is no friction. Don't worry about the mass just now.

The boy is sliding down the slope. What is the size of the force that is acting on him down the slope (this is where your force diagram comes in handy) .....
(Original post by TheManor)
just call the mass "m"
Ahh the mass cancels on each side and then you have 9.8sin(20)=a

Thank you so much!

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