# what does f'(x) actually mean?

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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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#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

So imagine the first equation is f(x) =

Thats the same as y=

So, f'(x)

Is dy/dx

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#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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#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)

If y=f(x) then dy/dx=f'(x)

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#10

(Original post by

thank you guys! i'm not exactly the best mathematician out there!

**stainless-style**)thank you guys! i'm not exactly the best mathematician out there!

Have you not come across it in class or in a textbook before?

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(Original post by

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?

**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?

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#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)

"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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#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.......

Thanks in advance

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.......

Thanks in advance

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#17

(Original post by

cheers mate, so it would be

Ac' = 8q - 3 - 0.25/q^2? ........i think

**Saiko21**)cheers mate, so it would be

Ac' = 8q - 3 - 0.25/q^2? ........i think

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#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

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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#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?

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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#20

(Original post by

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

**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

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