Query concerning derivative of e^x

[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.

 

[tiny-tim]

Hi Kaimyn!

(try using the X² tag just above the Reply box )

… is there a way to calculate NP derivatives such as: $$2^{x}$$?

2^x = (e^ln(2))^x = e^x.ln(2), so (2^x)' = ln(2).2^x

(and since ln(e) = 1, e is the only number with (e^x)' =e^x)

 

[slider142]

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 e^x with respect to x is e^x. 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)).

 

[HallsofIvy]

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.

 

[HallsofIvy]

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

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

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 a^x is a constant time a^x.

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
[tex]\lim_{h\rightarrow 0}\frac{a^h- 1}{h}= 1[/itex]
and we define e to be that number.

 

[Random Variable]

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.

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

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

 

[Kaimyn]

Thank you for all your helpful replies.

2^x = (e^ln(2))^x = e^x.ln(2), so (2^x)' = ln(2).2^x

Thanks, now it actually makes sense.

If f(x)= a^x, for any positive number a, then
[tex]\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".

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

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.