# The derivative of function under constraint

• Raghav Gupta
In summary, the conversation discusses a problem in which a function is given and the goal is to find its derivative. The function satisfies the condition f(x+y) = f(x)f(y) for all x and y, and it is known that f(5) = 2 and f'(0) = 3. The conversation goes through various attempts and hints in finding f'(5), ultimately concluding that it is equal to 6. However, there is a subtle error in the question as the given information is not consistent enough to determine the function completely.
Raghav Gupta

## Homework Statement

If f(x+y) = f(x)f(y) for all x and y and f(5) = 2, f'(0)=3, then f'(5) is
a) 5
b) 6
C) 0
d) None of these

## Homework Equations

Don't know which equation to apply. I was thinking of Rolle's Theorem and mean value theorem here but it is not helping.

## The Attempt at a Solution

Found f(0) = 1

Raghav Gupta said:

## Homework Statement

If f(x+y) = f(x)f(y) for all x and y and f(5) = 2, f'(0)=3, then f'(5) is
a) 5
b) 6
C) 0
d) None of these

## Homework Equations

Don't know which equation to apply. I was thinking of Rolle's Theorem and mean value theorem here but it is not helping.

You're not actually told that the function is differentiable anywhere other than at zero, so the first thing to do is prove that $f'(5)$ actually exists. Therefore you should start from $$f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}.$$

Notice also that, because f(x+ y)= f(x)f(y), f(x+ 0)= f(x)= f(x)f(0) so we must have f(0)= 1. That helps!

I think we're waiting for OP to respond.

HallsofIvy said:
Notice also that, because f(x+ y)= f(x)f(y), f(x+ 0)= f(x)= f(x)f(0) so we must have f(0)= 1. That helps!
But I have already shown that in my attempt of post 1. How it is helping me?
pasmith said:
$$f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h}.$$
Now what should I do?
I am getting a 0/0 form on right side and L Hopital rule also is not beneficial here.

Raghav Gupta said:
But I have already shown that in my attempt of post 1. How it is helping me?

Now what should I do?
I am getting a 0/0 form on right side and L Hopital rule also is not beneficial here.

Derivative limits always have a 0/0 form. Use f(x+y)=f(x)f(y).

Dick said:
Derivative limits always have a 0/0 form. Use f(x+y)=f(x)f(y).
So, I am getting,
$$f'(5) = \lim_{h \to 0} \frac{f(5)f( h) - f(5)}{h}.$$
Now f(5) = 2 and f(0)=1, so f(h) = 1
Again getting 0/0 form. What I am supposed to do here ?

Raghav Gupta said:
So, I am getting,
$$f'(5) = \lim_{h \to 0} \frac{f(5)f( h) - f(5)}{h}.$$
Now f(5) = 2 and f(0)=1, so f(h) = 1
Again getting 0/0 form. What I am supposed to do here ?

You are given f'(0)=3. What's that as a difference quotient? Halls gave you a hint in post 3.

$$f'(0) = \lim_{h \to 0} \frac{f(0)f( h) - f(0)}{h}.$$

Now what? Again 0/0 form.

Raghav Gupta said:
$$f'(0) = \lim_{h \to 0} \frac{f(0)f( h) - f(0)}{h}.$$

Now what? Again 0/0 form.

Think about it some more. The limit is 3. You are given that. As for what f(0) is you were also hinted that.

Okay , applying L Hopital rule getting f'(h) = 3
Then using in LH rule once again in f'(5) getting 2f'(h) = 6= f'(5)
Was it supposed to be done by LH rule?

No, you don't need LH. And what is f'(h)=3 supposed to mean? f'(0)=3. You haven't thought about this enough.

Dick said:
No, you don't need LH. And what is f'(h)=3 supposed to mean? f'(0)=3. You haven't thought about this enough.
Not,getting
$$f'(0) =3= \lim_{h \to 0} \frac{f( h) - 1}{h}.$$

Raghav Gupta said:
Not,getting
$$f'(0) =3= \lim_{h \to 0} \frac{f( h) - 1}{h}.$$

Good so far. Now figure out f'(5). It has that expression in it. Don't you see?

Dick said:
Good so far. Now figure out f'(5). It has that expression in it. Don't you see?
Yeah f'(5) = 2*3 = 6 by that expression. Thanks Dick.
Thanks to other people also.

There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.

pasmith said:
There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.
How f(x) = f(1)x ?
A counter example-
Putting x = 0
We get f(0) = f(1)
Putting x= 5
We get f(5) = f(1)
But f(5) ≠ f(0)
As we are given f(5) = 2 and we know f(0) = 1

Raghav Gupta said:
How f(x) = f(1)x ?
A counter example-
Putting x = 0
We get f(0) = f(1)
No, f(0) = (f(1))0 = 1
Raghav Gupta said:
Putting x= 5
We get f(5) = f(1)
But f(5) ≠ f(0)
As we are given f(5) = 2 and we know f(0) = 1

Oh, so it is
SammyS said:
No, f(0) = (f(1))0 = 1
(f(1))x not f(1)x
How can we prove f(x) = (f(1))x ?

Raghav Gupta said:
Oh, so it is

(f(1))x not f(1)x
How can we prove f(x) = (f(1))x ?
Ask Mr. (or Ms.) Smith to be certain.

...but

If f(x) = Cx for some C > 0, then we have f(x+y) = f(x)f(y) .

SammyS said:
...but

If f(x) = Cx for some C > 0, then we have f(x+y) = f(x)f(y) .
How? Is it some theorem and if it is what it is called?
Can you tell me that Mr. Sammy ?

Raghav Gupta said:
How? Is it some theorem and if it is what it is called?
Can you tell me that Mr. Sammy ?
Try it. Use some basic algebra.

If I write it out for you, then Mark and other mentors & homework helpers would be displeased with me. (In fact I would then be displeased with me. )

SammyS said:
Try it. Use some basic algebra.

If I write it out for you, then Mark and other mentors & homework helpers would be displeased with me. (In fact I would then be displeased with me. )
Got it, it was basic algebra really.
pasmith said:
There is a subtle error in the question.

You can prove that $f(x) = f(1)^x = e^{x \ln (f(1))}$ and differentiate that by the chain rule. The function is completely determined by f(1), but you're given two pieces of information with which to try to find it. They are not consistent.
Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))x
Differentiating,
f'(x) = (f(0))xlogef(0)
Putting x = 0
f'(0) = 0, which is not true since in question it is given f'(0) = 3

Raghav Gupta said:
Got it, it was basic algebra really.

Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))x
Differentiating,
f'(x) = (f(0))xlogef(0)
Putting x = 0
f'(0) = 0, which is not true since in question it is given f'(0) = 3
As pasmith said, the conditions given are inconsistent.

SammyS said:
As pasmith said, the conditions given are inconsistent.

So following this method the answer could be none of these?
But yes there is error in question as well.
(It is looking like my every question of test that I am asking today is resulting in an error except that matrix one)

Raghav Gupta said:
Got it, it was basic algebra really.

Wow, is this a paradox( This seems true but contradicts itself) ?
f(x) = (f(0))x

No, it's $f(x) = (f(1))^x$. As you have established, we must have $f(0) = 1$.

If $f(x + y) = f(x)f(y)$ then $f'(x) = f(x)f'(0)$ and $f(0) = 1$. This initial value problem has the solution $f(x) = e^{xf'(0)}$, so $f(1) = e^{f'(0)}$ and thus $f(x) = (f(1))^x$.

The inconsistency is that if $f'(0) = 3$ then $f(1) = e^3$, whilst if $f(5) = 2$ then $f(1) = 2^{1/5}$.

The question setter has obviously just exploited the fact that if $f(x + y) = f(x)f(y)$ then $f'(x) = f(x)f'(0)$ and has provided arbitrary values for $f(5)$ and $f'(0)$ without realizing that these values are not independent.

BvU and SammyS
pasmith said:
If $f(x + y) = f(x)f(y)$ then $f'(x) = f(x)f'(0)$ and $f(0) = 1$.
I think you have the prime in the wrong place.

If $\displaystyle f(x + y) = f(x)f(y)$ then $f'(x) = f'(x)f(0)$ and $f(0) = 1$

Never mind. Ray Vickson has pointed out how you were using this.

( I misinterpreted the and in the above .)

Last edited:
SammyS said:
I think you have the prime in the wrong place.

If $\displaystyle f(x + y) = f(x)f(y)$ then $f'(x) = f'(x)f(0)$ and $f(0) = 1$

The functional equation gives ##f(0) = f(0+0) = f(0)^2##, so ##f(0) = 0## or ##f(0) = 1##. We must have ##f(0) = 1##, because the other case ##f(0) = 0## would give ##f(x) = f(x+0) = f(x) f(0) = 0## for all ##x##, and that would contradict the nonzero value given to ##f'(0)##.

We thus have
$$\frac{f(x+h) - f(x)}{h} = \frac{ f(x) f(h) - f(x)}{h} = f(x) \frac{f(h)-1}{h} \\ = f(x) \frac{f(h) - f(0)}{h}$$
Now take ##h \to 0##.

pasmith said:
No, it's $f(x) = (f(1))^x$.
SammyS said:
If f(x) = Cx for some C > 0, then we have f(x+y) = f(x)f(y) .
Why f(x) is only equal to f((1))x?
As Sammy is saying it could be any constant.
So we can also write f(x) =(f(2))x ?
Ray Vickson said:
The functional equation gives ##f(0) = f(0+0) = f(0)^2##, so ##f(0) = 0## or ##f(0) = 1##. We must have ##f(0) = 1##, because the other case ##f(0) = 0## would give ##f(x) = f(x+0) = f(x) f(0) = 0## for all ##x##, and that would contradict the nonzero value given to ##f'(0)##.

We thus have
$$\frac{f(x+h) - f(x)}{h} = \frac{ f(x) f(h) - f(x)}{h} = f(x) \frac{f(h)-1}{h} \\ = f(x) \frac{f(h) - f(0)}{h}$$
Now take ##h \to 0##.
It's looking like I'm going round and round if you see my encounter with Dick in this thread.
So if we take
## h\to 0 ##
Then $$f'(x)= f(x) \frac{f(h) - f(0)}{h}$$
Now what to do?

Last edited:
I can also prove the inconsistency by taking,
f(x) = (f(5))x, it is given in question that f(5) = 2.
Now I am going to show that if f(5) = 2 then f'(0) ≠ 3
Differentiating,
## f'(x) = f(5)^x ln f(5) ##
Putting x = 0 and we know from question f(5) = 2
We see f'(0) = ln2 ≠ 3
So the values are dependent.
Is this also a way of correctly thinking?

Raghav Gupta said:
Why f(x) is only equal to f((1))x?
As Sammy is saying it could be any constant.
So we can also write f(x) =(f(2))x ?

It's looking like I'm going round and round if you see my encounter with Dick in this thread.
So if we take
## h\to 0 ##
Then $$f'(x)= f(x) \frac{f(h) - f(0)}{h}$$
Now what to do?

No: this equation is false. For finite ##h > 0## the right-hand-side is NOT equal to ##f'(x)##. Derivatives involve a limit, not just a difference-ratio.

Ray Vickson said:
No: this equation is false. For finite ##h > 0## the right-hand-side is NOT equal to ##f'(x)##. Derivatives involve a limit, not just a difference-ratio.
But if you see I have written ## h\to 0 ##

Raghav Gupta said:
But if you see I have written ## h\to 0 ##

No, you have not written it properly and in the correct place. You should have written
$$f'(x) = f(x) \, \lim_{h \to 0} \frac{f(h) -f(0)}{h}$$
THAT would be a correct statement.

Do you see how it is different from what you wrote? (This not mere quibbling; if you are not careful to write things correctly, they can come back later to cause you to make difficult-to-trace errors. I am trying to help you develop good habits that will be useful to you later, when problems are a lot more complicated.)

Raghav Gupta

## 1. What is the definition of a derivative under constraint?

A derivative under constraint refers to the rate of change of a function with respect to one of its variables, while taking into account a given constraint or limitation. This constraint can be in the form of an equation or condition that the function must satisfy.

## 2. How is the derivative of a function under constraint calculated?

The derivative of a function under constraint is calculated using the method of Lagrange multipliers, which involves finding the critical points of the function and the constraint equation, and then solving a system of equations to determine the optimal values.

## 3. What is the significance of the derivative under constraint in real-world applications?

The derivative under constraint is important in many real-world applications, such as optimization problems in economics, engineering, and physics. It allows us to find the maximum or minimum value of a function while satisfying a given constraint, which is useful in decision-making and problem-solving.

## 4. Can the derivative under constraint be negative?

Yes, the derivative under constraint can be negative. This indicates a decreasing rate of change of the function with respect to the constrained variable. It could also mean that the function is decreasing while satisfying the given constraint.

## 5. How does the derivative under constraint differ from the regular derivative?

The derivative under constraint takes into consideration a given constraint, while the regular derivative does not. This means that the derivative under constraint is calculated with an additional variable (the Lagrange multiplier) and involves solving a system of equations. The regular derivative, on the other hand, only involves finding the rate of change of a function with respect to a single variable.

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