# what does f'(x) actually mean?

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Thread starter 12 years ago
#1
i understand f(x) is the function of x. but when i get questions asking me to find f'(x) i don't understand what they mean?!
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12 years ago
#2
Differentiate f(x)
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12 years ago
#3
That's the first derivative of the equation
So imagine the first equation is f(x) =
Thats the same as y=
So, f'(x)
Is dy/dx
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12 years ago
#4
your need to differentiate f(x)
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12 years ago
#5
The derivative with respect to x.
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12 years ago
#6
f'(x) is usually taken to be the derivative of f(x), i.e. the function you get after differentiating f(x)
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12 years ago
#7
f'(x) is the first derivative

edit: very slow...
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12 years ago
#8
It's the derived function: differentiate whatever f(x) is and you've found f'(x).

If y=f(x) then dy/dx=f'(x)
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Thread starter 12 years ago
#9
thank you guys! i'm not exactly the best mathematician out there!
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12 years ago
#10
(Original post by stainless-style)
thank you guys! i'm not exactly the best mathematician out there!
Well tbh knowing that f'(x) means differentiate f(x) with respect to x isn't really a question of being a good mathematician or not!! The whole point is that first of all you should learn the notation, either from a book or because a teacher has introduced it to you, and then you tackle problems that use this notation.

Have you not come across it in class or in a textbook before?
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12 years ago
#11
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Thread starter 12 years ago
#12
(Original post by davros)
Well tbh knowing that f'(x) means differentiate f(x) with respect to x isn't really a question of being a good mathematician or not!! The whole point is that first of all you should learn the notation, either from a book or because a teacher has introduced it to you, and then you tackle problems that use this notation.

Have you not come across it in class or in a textbook before?
i've only seen it as dy/dx but atleast i'll know from now on!
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12 years ago
#13
f''(x) means differentiate f'(x). i.e. d2y/dx2

"The second derivative". f with two dashes :P Just differentiate f'(x)

(Just added that in as I thought it would be helpful)
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12 years ago
#14
Hey guys, i was wondering if someone could help me...

I have an average cost equation of:

AC= 4q^2 - 3q + 2 + 0.25/q

Can someone please help me work out the first and second derivative, the fraction in there is confusing me quite badly.......

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12 years ago
#15
Rewrite 0.25/q as 0.25q^-1, and differentiate as usual.
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12 years ago
#16
cheers mate, so it would be

Ac' = 8q - 3 - 0.25/q^2? ........i think
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12 years ago
#17
(Original post by Saiko21)
cheers mate, so it would be

Ac' = 8q - 3 - 0.25/q^2? ........i think
Yes, that's right.
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12 years ago
#18
ok another similar question lol. sorry bout this....

I have p=100-5.5q
I worked out q to be q = p/5.5+100

wat is the first derivative of Q??

thanx
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12 years ago
#19
f(x) is a function of x, but f'(x) is the derivative with respect to what seems to make sense. In terms of a-level f'(x) pretty much means dy/dx, but you should realise that d(f(x))/dx = f'(x) doesn't require a y.
As for the post, it depends what you are differentiating with respect to but I assume it is p (i.e. do you want dq / dp ?). Btw, q = (100 - p) / 5.5, then also q = 100/5.5 - p/5.5, now if I said q = y, and p = x, then y = 100/5.5 - x/5.5. Can you find dy/dx?
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12 years ago
#20
(Original post by Saiko21)
ok another similar question lol. sorry bout this....

I have p=100-5.5q
I worked out q to be q = p/5.5+100
Afraid not...
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