# 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.

## 1. What is the derivative of e^x?

The derivative of e^x is simply e^x itself. This means that the slope of the curve at any point on the graph of e^x is equal to the y-value of that point.

## 2. How do you find the derivative of e^x?

To find the derivative of e^x, you can use the power rule of differentiation. This involves taking the exponent of e (which is x) and multiplying it by the coefficient of e (which is also x). So, the derivative of e^x is e^x.

## 3. Why is the derivative of e^x equal to e^x?

This is because the function e^x is its own derivative. In other words, when you take the derivative of e^x, you are essentially finding the slope of the function at any given point, and since e^x has a constant slope of e^x, its derivative remains e^x.

## 4. What is the significance of e^x in calculus?

e^x is a very important function in calculus as it is the base of the natural logarithm. Its derivative is also used in many applications, such as in economics, physics, and engineering, to describe exponential growth and decay.

## 5. Can the derivative of e^x be negative?

No, the derivative of e^x cannot be negative. This is because the derivative of e^x is always e^x, which is a positive value. However, the slope of the tangent line to the graph of e^x can be negative at certain points, but this is not the same as the actual derivative.

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