Derivative of itself = itself

Calculus Level 2

How many function(s) $f(x)$ satisfies the condition

$f(x) = f ' (x)$

where $f ' (x)$ means the derivative of $f(x)$ respect to $x$ ?

If you think the answer is infinitely many, submit your answer as -1.

The answer is -1.

2 solutions

Md Zuhair
Feb 21, 2017

Okay, So this can be mathematically proved as below,

IF $f(x)=y$ (Say) And $f'(x) = \dfrac{dy}{dx}$ . So,

By our equation $y= \dfrac{dy}{dx}$

Hence $dx = \dfrac{dy}{y}$

hence INTEGRATING EACH SIDE WE GET

$x = lny +c$ Where c is the constant of integration

Hence $x-c = lny$

Or $e^{x-c} = y$

or, $y= \boxed{e^{x-c}}$ .

Hence there are infinitely many solutions for any $c \epsilon R$

@Christian Daang , Please Remove this answer box , and keep a MCQ. Because anyone can 1st put 1 can check as he has 2 chances left. If you like, you may do it.

Md Zuhair - 4 years, 3 months ago

I try to edit the question, I think, it is okay now. :)

Christian Daang - 4 years, 3 months ago

Formally, this proves "If $y=f(x) > 0$ is a solution to $f'(x) = f(x)$ , then $y=e^{x-c}$ for some $c\in \mathbb{R}$ ."

In contrast, the proof [and it is a proof] that there are infinitely many such functions is one line: $\forall c\in \mathbb{R} : \frac{d}{dx}\left[e^{x-c}\right] = e^{x-c}.$

You should be comfortable calling both "proof"

Brian Moehring - 4 years, 3 months ago

Ya, but I deduced the equation $e^{x-c}$ from the given equation where as he just gave us the equation. So i call it a complete one.

Md Zuhair - 4 years, 3 months ago

You need to be careful. I feel like in your quest for some idea of "proof", you aren't looking at what you've technically shown...

You didn't technically show all the functions of the form $f(x) = e^{x-c}$ are solutions. You only showed that if $f(x)$ is a positive solution, then it has that form for some real $c$ . (given the form of your proof, it would still be possible for there to be no solutions!!)

To prove that all of those functions are solutions, you would need to take the derivative and see it gives itself. Then to show you have infinitely many solution, you need to show that $\left\{e^{x-c} : c\in \mathbb{R}\right\}$ is an infinite set.

Brian Moehring - 4 years, 3 months ago

@Brian Moehring – Ok sir, Please cool down. Okay. So i Agree my mistake. Thank you. Very Sorry.

Md Zuhair - 4 years, 3 months ago

@Md Zuhair – Oh, I never was mad (emotion is one of those things that doesn't come through very well in text), and there's no reason to apologize.

I simply see a lot of very bright students who refuse to check whether they've proven what they intended to prove. If nothing else, you should just take my comments as someone on the internet claiming that learning to critique your own proofs is an incredibly valuable skill, and that it's usually as profitable as writing the proof itself.

Brian Moehring - 4 years, 3 months ago

@Brian Moehring – Ya sir, Sure. I will keep it in mind. And i will accept all kinds of comments on my solutions. Thank you.

Md Zuhair - 4 years, 3 months ago

Christian Daang
Feb 21, 2017

The possible example of function $f(x)$ is $e^{x + k}$ where k $\in \mathbb{R}$ which means there are infinitely many functions that satisfy the condition.

FYI I've made the infinite case yield a value of -1, so that people who think there is a unique solution will realize that they made a mistake (instead of assuming that they are correct).

Calvin Lin Staff - 4 years, 3 months ago

It still says 1 in the original problem (which is what I put since there are infinite solutions).

Sahil P. - 4 years, 3 months ago

Ah yes, I saved over my edit. Those who answered 1 have been marked correct.

Calvin Lin Staff - 4 years, 3 months ago

Derivative of itself = itself

The answer is -1.

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