# Why is $e^x$ the only function that is its own derivative? [duplicate]

I've heard that $f(x) = Ae^x$ is only function (both elementary and non-elementary) that satisfies the property $f(x)=\frac{df(x)}{dx}$. Is this true (and if it's true, is there a definitive way to prove it)?

## marked as duplicate by Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 21 '18 at 5:59

• This isn't true, but it's very close to true. $f(x) = A e^x$ works for any real $A$. Taking $A=1$ or $A=0$ gives the two examples you gave in your question. The appropriate proof for this fact depends on how you've defined the exponential function - some authors define $e^x$ as the solution to your property which satisfies $f(0) = 1$. – B. Mehta Feb 20 '18 at 19:40

Let's do a "function research" for functions of the type $f(x) = f^{\prime}(x)$.

From $f(x) = f^{\prime}(x)$ it follows that $f$ should be infinitely often differentiable.

So, you may try to check whether there is a meaningful Taylor series around $x= 0$ describing such an $f$. So set $A = f(0)$ and let's try $A \neq 0$. So, you get

$$\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n = A \sum_{n=0}^{\infty}\frac{1}{n!}x^n = A \cdot E(x)$$.

Now you study $E(x)$ and find all the nice properties like $E(x+y) = E(x)E(y)$ etc. and realize that this function is uniquely determined and is usally written as $e^x$.

Any constant times $e^x$ also has this property.

To see that these are the only examples, suppose we have a function $g(x)$ with $g'(x)=g(x)$. Let $h(x)=g(x)e^{-x}$. Note that $e^x$ is never $0$ so $h(x)$ is well-defined. We compute $$h'(x)=g'(x)e^{-x}-g(x)e^{-x}=0\implies h(x)=\text {constant}$$

and we are done.

• Very nice. +1${}{}{}{}{}{}{}{}{}$ – DonAntonio Feb 20 '18 at 19:44
• While I love this proof, OP should note that this assumes we know $e^x$ is its own derivative, and shows that it is the only function which is its own derivative (up to constant multiplication). – B. Mehta Feb 20 '18 at 19:44
• @B.Mehta : your concern is addressed in this answer math.stackexchange.com/a/1292586/72031 – Paramanand Singh Feb 21 '18 at 14:11

Let $y=f(x)$ thus the condition

$$f’(x)=f(x)$$

is a differential equation and for the Theorem of Existence and Uniqueness an unique solution exists up to a constant.

Notably by separation of variables

$$\frac{dy}{dx}=y\implies \frac{dy}{y}=dx\implies \log y = x+c$$

Which shows that the unique solution is the inverse of $\log$ function.