# Query concerning derivative of e^x

• Kaimyn
In summary, e is the unique number such that e^x is equal to its derivative. It can be found through the limit of the derivative of a^x as a approaches 1. The derivative of e^x can also be calculated using the chain rule.

#### Kaimyn

I've been studying calculus and have always been confused about the property of e^x.
"e is the unique number such that e^x is equal to its derivative."

I haven't ever really understood why, but have figured it would pop up some time later. Unfortunately, it still hasn't popped up, so I've decided to ask. How is e^x the derivate of itself?

I was mucking around with numbers a bit and came up with a couple of things:

Asumming e^x is its derivative:
$$y = e^{x^{2}+2}$$
Then:
$$\frac{dy}{dx} = e^{x^{2}+2}$$
However (using the chain rule):
$$\frac{dy}{dx} = 2xe^{x^{2}+2}$$
Which isn't the same thing...

Also, while I'm on the topic, is there a way to calculate NP derivatives such as: $$2^{x}$$?

I am sorry if this seems basic, or is commonly asked, but I just can't seem to figure it out.

Hi Kaimyn! (try using the X2 tag just above the Reply box )
Kaimyn said:
… is there a way to calculate NP derivatives such as: $$2^{x}$$?

2x = (eln(2))x = ex.ln(2), so (2x)' = ln(2).2x (and since ln(e) = 1, e is the only number with (ex)' =ex)

Kaimyn said:
Asumming e^x is its derivative:
$$y = e^{x^{2}+2}$$
Then:
$$\frac{dy}{dx} = e^{x^{2}+2}$$
However (using the chain rule):
$$\frac{dy}{dx} = 2xe^{x^{2}+2}$$
Which isn't the same thing...
The derivative of ex with respect to x is ex. If x is a function of t (ie., x = t2) and you want the derivative with respect to t of the whole expression, then of course you have to use the chain rule.

Also, while I'm on the topic, is there a way to calculate NP derivatives such as: $$2^{x}$$?
One common method is to take a logarithm (it doesn't have to be the natural logarithm) of both sides and use implicit differentiation (ie., use the fact that if A(x) = B(x) and A is differentiable, then A'(x) = B'(x)).

Kaimyn said:
I've been studying calculus and have always been confused about the property of e^x.
"e is the unique number such that e^x is equal to its derivative."

I haven't ever really understood why, but have figured it would pop up some time later. Unfortunately, it still hasn't popped up, so I've decided to ask. How is e^x the derivate of itself?

I was mucking around with numbers a bit and came up with a couple of things:

Asumming e^x is its derivative:
$$y = e^{x^{2}+2}$$
Then:
$$\frac{dy}{dx} = e^{x^{2}+2}$$[/itex]
You just said ""e is the unique number such that e^x is equal to its derivative."
Given that, it is impossible that $$e^{x^2}$$ is also its own derivative!
Why would you assume that?

However (using the chain rule):
$$\frac{dy}{dx} = 2xe^{x^{2}+2}$$
Which isn't the same thing...

Also, while I'm on the topic, is there a way to calculate NP derivatives such as: $$2^{x}$$?

I am sorry if this seems basic, or is commonly asked, but I just can't seem to figure it out.

Both tiny-tim and slider showed how to get the derivative of ex from the derivative of ln(x). Here's how to get it directly.

If f(x)= ax, for any positive number a, then
$$\frac{f(x+h)- f(x)}{h}=\frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h}[/itex] Thus, [tex]\frac{da^x}{dx}= a^x\lim_{h\rightarrow 0}\frac{a^h- 1}{h}$$

In other words, the dependence on x can be taken out of the limit: since that limit does not depend on x, it is a constant.

So, for any positive real number, a, the derivaive of ax is a constant time ax.

It is not too difficult to see that if a= 2, say, then that limit must be less than 1 or that if a= 3, it is larger than 1. Thus, there exist a value of a between 2 and 3 such that
$$\lim_{h\rightarrow 0}\frac{a^h- 1}{h}= 1[/itex] and we define e to be that number. Last edited by a moderator: Assuming you can find the derivative of ln(x) straight from the defintion, you can use the chain rule to find the derivative of e^x. [tex] \frac{d(ln(e^x))}{dx} = \frac{d(x)}{dx} = 1$$

but $$\frac{d(ln(e^x))}{dx} = \frac{d(ln(u))}{du}\frac{d(e^x)}{dx} = \frac{1}{e^x}\frac{d(e^x)}{dx}$$

therefore, $$\frac{1}{e^x}\frac{d(e^x)}{dx} = 1$$

or $$\frac{d(e^x)}{dx} = e^x$$

tiny-tim said:
2x = (eln(2))x = ex.ln(2), so (2x)' = ln(2).2x

Thanks, now it actually makes sense.

HallsofIvy said:
If f(x)= ax, for any positive number a, then
$$\frac{f(x+h)- f(x)}{h}=\frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\frac{a^h- 1}{h}[/itex] Now I have "proof". Random Variable said: [tex] \frac{d(ln(e^x))}{dx} = \frac{d(x)}{dx} = 1$$

but $$\frac{d(ln(e^x))}{dx} = \frac{d(ln(u))}{du}\frac{d(e^x)}{dx} = \frac{1}{e^x}\frac{d(e^x)}{dx}$$

therefore, $$\frac{1}{e^x}\frac{d(e^x)}{dx} = 1$$

or $$\frac{d(e^x)}{dx} = e^x$$

Another mathematical proof that makes sense.

Thanks again for all the helpful replies.