Derivative of e^x

[applejuicefool]

Hi, I'm new here!

I am going to be taking a Cal II course this summer, so I was over on Khan Academy boning up on my Calculus, when I encountered what seems to be to be an odd discrepancy. I am positive it is just an artifact of my ignorance, so I am hoping someone here can explain it to me.

So I know that the derivative of e^x is supposed to be simply e^x. Therefore, it seems to me that the derivative of e^2 should be simply e^2.

The website wolframalpha.com disagrees, however. It claims that the derivative of e^2 is 0, while e^2 is a transcendental number 7.389056..... etc.

Can anyone explain this quandary to me?

 

[Deleted member 4993]

Hi, I'm new here!

I am going to be taking a Cal II course this summer, so I was over on Khan Academy boning up on my Calculus, when I encountered what seems to be to be an odd discrepancy. I am positive it is just an artifact of my ignorance, so I am hoping someone here can explain it to me.

So I know that the derivative of e^x is supposed to be simply e^x. Therefore, it seems to me that the derivative of e^2 should be simply e^2.

The website wolframalpha.com disagrees, however. It claims that the derivative of e^2 is 0, while e^2 is a transcendental number 7.389056..... etc.

Can anyone explain this quandary to me?

e^x is a function of x. For a given different values of value of x - you would get different values of 'x'. So e^x is a variable - it is a dependent variable (dependent on x).

However e² (= 2.718282...*2.718282...) is a constant like 2² or π² or 10^1/3

e² is irrational and it has an approximate value of 7.3891... (e = 2.718282).

Thus, like every constant $\displaystyle \displaystyle{\frac{d}{dx}e^2 \ = \ 0}$

 

[applejuicefool]

e^x is a function of x. For a given different values of value of x - you would get different values of 'x'. So e^x is a variable - it is a dependent variable (dependent on x).

However e² (= 2.718282...*2.718282...) is a constant like 2² or π² or 10^1/3

e² is irrational and it has an approximate value of 7.3891... (e = 2.718282).

Thus, like every constant $\displaystyle \displaystyle{\frac{d}{dx}e^2 \ = \ 0}$

So how can it be said that $\displaystyle \displaystyle{\frac{d}{dx}e^x \ = \ e^x}$, if when you plug a value into x, the two sides aren't equal?

 

[Deleted member 4993]

So how can it be said that $\displaystyle \displaystyle{\frac{d}{dx}e^x \ = \ e^x}$, if when you plug a value into x, the two sides aren't equal?

Do you agree:

$\displaystyle \displaystyle{\frac{d}{dx}C \ = \ 0}$

 

[applejuicefool]

Do you agree:

$\displaystyle \displaystyle{\frac{d}{dx}C \ = \ 0}$

Yes. I understand what you're saying...e^2 is a constant, and derivatives of constants are 0. What I'm asking is how can it be said that derivative of e^x equals e^x, when they don't come out to be equal to each other when you plug values into x?

 

[pka]

Yes. I understand what you're saying...e^2 is a constant, and derivatives of constants are 0. What I'm asking is how can it be said that derivative of e^x equals e^x, when they don't come out to be equal to each other when you plug values into x?

Do you agree that $\displaystyle {e^x} \ne {e^2}$? Then $\displaystyle {D_x}({e^x}) \ne {D_x}({e^2})$.

 

[Steven G]

So how can it be said that $\displaystyle \displaystyle{\frac{d}{dx}e^x \ = \ e^x}$, if when you plug a value into x, the two sides aren't equal?

Let us move away from e^2.
Just consider the function x. We know that the derivative of x is 1.
Now let x=17. You want to say that 0=1.
You do not plug in the x value until AFTER you evaluate the derivative.
For example: f(x) = x^2. Find the derivative when x=5.

1st we find the derivative and then we plug in 5 for x. f ' (x) =2x and f' (5) = 2*5=10.

If you plug in the 5 1st then you will get 5^2 =25. And the derivative of 25 is 0.

Given any f(x) when you plug in a number you will get back a number and the derivative of that number will always be 0. For this reason we do not usually plug in the number first.

One last point. In my above example you are being asked to find f'(5) and obviously you'll need f' (x) to do this. You are being asked to find the derivative of f(5)!!

 

[applejuicefool]

Let us move away from e^2.
Just consider the function x. We know that the derivative of x is 1.
Now let x=17. You want to say that 0=1.
You do not plug in the x value until AFTER you evaluate the derivative.
For example: f(x) = x^2. Find the derivative when x=5.

1st we find the derivative and then we plug in 5 for x. f ' (x) =2x and f' (5) = 2*5=10.

If you plug in the 5 1st then you will get 5^2 =25. And the derivative of 25 is 0.

Given any f(x) when you plug in a number you will get back a number and the derivative of that number will always be 0. For this reason we do not usually plug in the number first.

One last point. In my above example you are being asked to find f'(5) and obviously you'll need f' (x) to do this. You are being asked to find the derivative of f(5)!!

Ok, here's what I was missing. The "x" in e^x is not simply a variable that you can plug whatever random value you want into. It is x as it relates to the function as a whole. So if you *do* just plug a random value into x, it no longer relates to the entire function, just to that particular value. If you say f(x) = e^x, and then set x to 2, then you are no longer talking about f(x) but f(2). f'(2) is not equivalent to f'(x).

Thanks for the help, guys!

 

[Steven G]

Ok, here's what I was missing. The "x" in e^x is not simply a variable that you can plug whatever random value you want into. It is x as it relates to the function as a whole. So if you *do* just plug a random value into x, it no longer relates to the entire function, just to that particular value. If you say f(x) = e^x, and then set x to 2, then you are no longer talking about f(x) but f(2). f'(2) is not equivalent to f'(x).

Thanks for the help, guys!

if you are asked to find f(7) the first thing you should look for is f(x). Once you find it then you plug 7 in for x.
If you are asked to the derivative of f(x) at x=8 they you are being asked to f'(x) when x=9. So first find f'(x) and plug in 8 for x

 

[Steven G]

I re-read your posts and think that I understand where you are confused.

You agree that (x^2)' =2x

But you think if we let x=3 that (3^2)' should equal 2*3

Here is the trouble (look at what is in red). On the rhs we plugged in 3 for x in 2x. AND 2x is the derivative of x^2.

On the lhs we did NOT plug in 3 for x into the derivative of x^2. Rather we plugged in 3 for x into x*2 (and then took the derivative). Since we did different things to both sides of the equation we should not expect to get the same results.

 

[stapel]

Since we did different things to both sides of the equation we should not expect to get the same results.

Well-said! :wink: