Derivatives: Solving F'(x) = [2f(x)+g(x)]' Problem

[gracy]

Homework Statement

$F'(x)$=$[2f(x)+g(x)]'$
$F'(0)=$
Given g'(0)=2 and f'(0)=5

Homework Equations

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?

 

[Mark44]

Homework Statement

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

Homework Equations

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 out?

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.

 

[epenguin]

What is the problem? None is stated.

 

[gracy]

What is the problem? None is stated.

But I want to know why?Which rule governs this action of taking 2 out?

 

[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.

 

[gracy]

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 your know what the problem is be we don't.

Sorry.I have edited the OP.

 

[haruspex]

Sorry.I have edited the OP.

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)

 

[Herman Trivilino]

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)

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

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

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?

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

 

[epenguin]

Homework Statement

$F'(x)$=$[2f(x)+g(x)]'$
$F'(0)=$
Given g'(0)=2 and f'(0)=5

Homework Equations

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?

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.)

 

[SteamKing]

Homework Statement

$F'(x)$=$[2f(x)+g(x)]'$
$F'(0)=$
Given g'(0)=2 and f'(0)=5

Homework Equations

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?

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'.

 

[Mark44]

Homework Statement

$F'(x)$=$[2f(x)+g(x)]'$
$F'(0)=$
Given g'(0)=2 and f'(0)=5

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.

 

[epenguin]

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

 

[gracy]

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

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

Right?

 

[Mark44]

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

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

Right?

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.

Your original equation:
$F'(x) =[2f(x)+g(x)]'$

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)]' ?

 

[gracy]

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

 

[haruspex]

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

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.

 

[gracy]

f'(x) means the value of that function when given the input (or 'evaluated at') x.

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

 

[haruspex]

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

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.

 

[gracy]

f'(x) and [f(x)]' are two ways of writing the same thing

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

 

[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.

 

[gracy]

The value of the derivative at x=2 is f '(2) = 3(2)^2=12.

derivative of what?

 

[vela]

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.

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

 

[Fredrik]

f'(x) and [f(x)]' are two ways of writing the same thing.

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.

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

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 $$F'=(2f+g)'.$$

 

[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.

 

[ehild]

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

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

 

[gracy]

f is a differentiable function. Its derivative is a function

Could anyone please tell me it's derivative is function of what?

 

[Fredrik]

Could anyone please tell me it's derivative is function of what?

f' is not a function of anything. It's just a function. The distinction between "is a function" and "is a function of" is never explained in books for some reason. I have always found that pretty annoying. The things that are "functions of" something are typically not functions.

Symbols like x,y and f are called variables. A more complicated string of text like f(x) is called an expression. The thing that's represented by a variable or an expression is called the value of the variable or the expression. When we say that x is a real number, what we really mean is that the value of x is a real number, i.e. that the variable x represents a real number.

Let x and y be real numbers such that x+y=1. This statement doesn't assign values to the variables x and y. Instead it prevents you from assigning arbitrary values to them. If you now assign the value 3 to x, then you have to assign the value -2 to y. The value of y is completely determined by the value of x. When this is the case, y is said to be a function of x. In this particular example, it's also the case that x is a function of y. However, neither x nor y are functions. They are real numbers.

A function should be thought of as a rule that associates exactly one element of a set Y with each element of a set X. That doesn't explain what a function is, but you don't need to know what functions are. You just need to know how to think about them, how to define them and how to tell if two given functions are the same. (If you want to know what a function is in a branch of mathematics based on set theory, you can take a look at this post. But don't worry if you don't understand it). To define a function f, you need to specify a set D and what f(t) is for all t in D. The set D is called the domain of f. To find out if two given functions f and g are the same, you must check if they have the same domain, and if they do, check if f(t)=g(t) for all t in that domain.

When a variable or an expression is a "function of" a variable, you can always use that to define an actual function. For example. If y is a function of x, then there's a unique function f such that regardless of what value we assign to x, we will have y=f(x). For example: Let x and y be real numbers such that x+y=1. Let f be the function defined by f(t)=1-t for all real numbers t. Now no matter what value we assign to x, we have y=1-x=f(x). Note that since f is a function, f(x) is a function of x.

In this example: f is a function. x,y and f(x) are real numbers. y and f(x) are functions of x.

 

[gracy]

Let f be the function defined by f(t)=1-t for all real numbers t.

Is "t"a number?

 

[gracy]

Let x and y be real numbers such that x+y=1. This statement doesn't assign values to the variables x and y. Instead it prevents you from assigning arbitrary values to them. If you now assign the value 3 to x, then you have to assign the value -2 to y. The value of y is completely determined by the value of x. When this is the case, y is said to be a function of x. In this particular example, it's also the case that x is a function of y. However, neither x nor y are functions.

Here x&y are functions as f'(x) is function,so x and y are just functions not functions of anything,right?

 

[Fredrik]

Is "t"a number?

That's a surprisingly difficult question to answer. When I'm looking at the expression 1-t or the equation f(t)=1-t, I'm inclined to say that t is a number (because the minus sign refers to subtraction of real numbers, and f only takes real numbers as input). But when I'm looking at the complete sentence, I'm inclined to say that since the variable t is never assigned a value, it would be weird to describe it as anything more than a variable.