# Homework Help: A derivative problem

1. Oct 16, 2015

### gracy

1. The problem statement, all variables and given/known data
$F'(x)$=$[2f(x)+g(x)]'$
$F'(0)=$
Given g'(0)=2 and f'(0)=5

2. Relevant equations

3. The attempt at a solution
I can solve this if the questions is as follows
$F'(x)$=$[f(x)+g(x)]'$
by applying sum rule
but I don't know what to do about 2?$F'(x)$=$[2f(x)+g(x)]'$
I tried to solve this as follows
$F'(x)$=$[2f(x)+g(x)]'$
=$[2.f'(x)+f(x).f'(2)]$ +$g'(x)]$ (by applying multiplication rule) here f prime denotes derivative of
=$[2.f'(x)+f(x)$+$g'(x)$]
But it is wrong . Actually I do know that 2 will be taken out (it has been given in my book as solution of this problem),But I want to know why?Which rule governs this action of taking 2 out?

Last edited: Oct 16, 2015
2. Oct 16, 2015

### Staff: Mentor

You're making this much more difficult than it actually is.
If f'(x) = g'(x), then f(x) = g(x) + C
IOW, if the derivatives of two functions are equal, then the two functions differ only by a constant.

3. Oct 16, 2015

### epenguin

What is the problem? None is stated.

Last edited: Oct 16, 2015
4. Oct 16, 2015

### gracy

5. Oct 16, 2015

### epenguin

Taking 2 out of what?

You can't just write a formula and that is a question. It has to have some form like given ... prove that.... where this, where that... Find the other...or something.
Best to quote the passage where this cones from, else you know what the problem is but we don't.

Last edited: Oct 16, 2015
6. Oct 16, 2015

### gracy

Sorry.I have edited the OP.

7. Oct 16, 2015

### haruspex

Your error is here:
(2f(x))' = f'(2)f(x)+2f'(x)
It should be
(2f(x))' = 2'f(x)+2f'(x)=(0)f(x)+2f'(x)=2f'(x)

8. Oct 16, 2015

### Mister T

The only relevant multiplication rule is that $[2f(x)]'=2f'(x)$.

So, $[2f(x)+g(x)]'=2f'(x)+g'(x)$.

The product rule tells us that

$[2f(x)]'=2f'(x)+(2)'f(x)=2f'(x)+0 \cdot f(x)=2f'(x)+0=2f'(x)$.

Note that $f'(2) \neq [f(2)]' \neq (2)'$.

9. Oct 16, 2015

### epenguin

So I guess the question is: F'(x) = ?

I think I see what your problem is - quite a common one still with students more advanced than you.

It looks like your have this addition formula for [ f(x) + g(x) ]' and you are regarding f and g as something rigidly fixed for all time. Or once chosen, then fixed. They aren't - nor is x for that matter. If the question had been find [ 2φ(x) + ψ(x) ]' you could probably have done it, if we want to do it very laboriously we could do it for that and then say "Let φ be f and ψ be g,..."

Students get that x, for instance can represent anything, they do not always get the way in use in algebra etc. a moment later it can represent anything else. So I can represent just any function by f. But I can represent just any function if I want by 2f. (I might have some reason for doing that, maybe I know that later in some calculation I'm doing that later on its going to get halved and it might give me a nicer expression, or maybe it has come out of some preceding calculation. But I'm free to do it for whatever reason or none.)

Last edited: Oct 16, 2015
10. Oct 16, 2015

### SteamKing

Staff Emeritus
There are rules which govern taking the derivative when functions are multiplied by constants, when functions are added or subtracted, multiplied or divided.

This article here shows what to do:

https://www.mathsisfun.com/calculus/derivatives-rules.html

Scroll down to the Table which reads 'Rules'.

11. Oct 16, 2015

### Staff: Mentor

Even in your edited post, it's still not clear what the question is. When you post a problem, write the problem statement clearly, so that there is no uncertainty in what it is you're trying to do.

Apparently, what you're supposed to do is find F'(0), given $F'(x)$=$[2f(x)+g(x)]'$, and g'(0) = 2, and f'(0) = 5. Using the basic rules of differentiation in the link that SteamKing gave, this is a straightforward problem.

12. Oct 16, 2015

### epenguin

Or maybe you did have a little difficulty with derivation, but the question is about functions of 0. This is just substitution.

13. Oct 17, 2015

### gracy

$f'(x)$=$[f(x)]'$=$(x)'$

here $(x)'$ means derivative of x.
but $f'(2)$≠$[f(2)]'$≠$(2)'$

Right?

14. Oct 17, 2015

### Staff: Mentor

f'(x) and [f(x)]' are two ways of writing the same thing. Neither is equal to x' in general.
We are 14 posts into what is a very straightforward problem. From other posts of yours that I have seen, you have a tendency relatively simple questions into threads that go on for 50 or more posts. Let's try not to let that happen here.

$F'(x) =[2f(x)+g(x)]'$

The right side means the derivative, with respect to x, of the quantity in the brackets. Using some of the first rules your learn when differentiation is presented in calculus, how can you rewrite [2f(x)+g(x)]' ?

15. Oct 17, 2015

### gracy

I have understood my OP.Here OP is original problem.But I am struggling with what
mister T wrote.Last line of #8

16. Oct 17, 2015

### haruspex

No. The derivative of f is the derivaitive of f. What makes you think it can morph into a derivative of x?
Maybe you don't understand the f(x) notation. It means the value returned by the function f when given the input x. The derivative of f, f', is another function. f'(x) means the value of that function when given the input (or 'evaluated at') x.

17. Oct 17, 2015

### gracy

Should not it be f'(x) means the derivative of that function when given the input (or 'evaluated at') x.

18. Oct 17, 2015

### haruspex

f(2) is the value of the function f given the input 2. That is simply a constant number, so does not change as x changes.
Part of your confusion comes from the fact that the prime notation, ', is a shorthand for a derivative with respect to some variable. You have to lnow from context which variable. In f'(x), it is obvious the variable is x. But in f'(2) it is no longer obvious. If it still means d/dx then f'(2) must be zero, since f(2) does not change when x changes.

19. Oct 17, 2015

### gracy

Then why not
$f'(2)$=$[f(2)]'$

20. Oct 17, 2015

### ehild

We do the operation in the brackets first. f '(2) is the value of the derivative at x=2. [f(2)] ' is the derivative of the number f(2), which is zero.

For example, let be f(x)=x^3. Its derivative is f '(x) = 3x^2.
The value of the derivative at x=2 is f '(2) = 3(2)^2=12.
On the other hand, [f(2)] ' = (2^3)' =0 as 2^3 is a constant.

21. Oct 17, 2015

### gracy

derivative of what?

22. Oct 17, 2015

### vela

Staff Emeritus
Except that f'(2) is the notation used to denote the derivative of f evaluated at x=2, which is generally not 0.

23. Oct 17, 2015

### Fredrik

Staff Emeritus
Yes, they both mean $\frac{d}{dx}f(x)$. I would add that [f(x)]' is a terrible confusing notation that should never be used.

Because the left-hand side is a good notation for the number you get when the function f' takes 2 as input, and the right-hand side is a terrible notation for either the function you get when you take the derivative of the function that takes every real number to f(2), or the number you get as output when that function takes some input (typically x) that isn't specified by the notation. So the right-hand side is either the number 0 or the function that takes every real number to 0.

f is a differentiable function. Its derivative is a function. It's denoted by f'. x is a number. f(x) is a number called "the value of f at x". Similarly, f'(x) is the value of f' at x. The notation [f(x)]' is absurd, because here the prime is on an expression that represents a number, and the derivative of a number is not defined.

The notation $$\frac{d}{dx}\, \text{some expression that involves x}$$ should be interpreted as "g'(x), where g is the function that takes an arbitrary real number x to the number represented by that expression". For example, $\frac{d}{dx} x^2$ denotes g'(x), where g is the function defined by $g(t)=t^2$ for all real numbers $t$.

A good way to write the first equality in post #1 is F'(x)=2f'(x)+g'(x). The right-hand side can be rewritten using the following definitions of the sum of two functions, and the product of a number and a function:

f+g is the function defined by (f+g)(x)=f(x)+g(x) for all real numbers x
kf is the function defined by (kf)(x)=k(f(x)) for all real numbers x

It's very easy to prove that (f+g)'=f'+g' and (kf)'=kf'. For all real numbers x, we have
$$F'(x)=2f'(x)+g'(x)=(2f)'(x)+g'(x)=((2f)'+g')(x) =(2f+g)'(x).$$ This is equivalent to $$F'=(2f+g)'.$$

24. Oct 17, 2015

### gracy

Thanks @Fredrik .It's a" big "/detailed answer.I will try to understand each and every line.It will take time.Calculus is very new to me.

25. Oct 17, 2015

### ehild

If you want to be very precise it can be written as f'(x) |x=2, but it is long and inconvenient.