# Derivative of number e

#### wishmaster

##### Active member
I have to derivate this function:

$$\displaystyle x^3-5x^2+4x-e^x$$ where $x$ on e is actualy $x^0$,but i dont know how to write it here.....
i know how to derivate other terms,just this e is suspicious,because its a complement of the function if i am right.....

Another derivate i have to calculate is:

$$\displaystyle ln(x+\sqrt{1+x^{-25}}$$

Thank you all for the help!

#### MarkFL

Staff member
Just so we're clear, you are given:

$$\displaystyle f(x)=x^3-5x^2+4x-e^{x^0}$$

Right?

Note: In $\LaTeX$, to create an exponent having more than 1 character, enclose that exponent in curly braces, in the above I used the code e^{x^0}.

#### wishmaster

##### Active member
Just so we're clear, you are given:

$$\displaystyle f(x)=x^3-5x^2+4x-e^{x^0}$$

Right?

Note: In $\LaTeX$, to create an exponent having more than 1 character, enclose that exponent in curly braces, in the above I used the code e^{x^0}.

Yes,you are right! So i have learned something new

#### MarkFL

Staff member
Can you simplify $e^{x^0}$?

Staff member

#### MarkFL

Staff member
So then $x$ is zero,and $e$ is also $0$ ?
No, $e$ is a transcendental constant (like $\pi$), but what I was getting at, can we say:

$$\displaystyle e^{x^0}=e$$

when $x=0$ ?

In other words, can we state $0^0=1$ ?

#### wishmaster

##### Active member
No, $e$ is a transcendental constant (like $\pi$), but what I was getting at, can we say:

$$\displaystyle e^{x^0}=e$$

when $x=0$ ?

In other words, can we state $0^0=1$ ?
So then remains just $e$ when i derivate?

#### MarkFL

Staff member
So then remains just $e$ when i derivate?
No, apply the exponential and chain rules:

$$\displaystyle \frac{d}{dx}\left(e^{f(x)} \right)=e^{f(x)}\frac{d}{dx}\left(f(x) \right)$$

What do you find?

#### wishmaster

##### Active member
No, apply the exponential and chain rules:

$$\displaystyle \frac{d}{dx}\left(e^{f(x)} \right)=e^{f(x)}\frac{d}{dx}\left(f(x) \right)$$

What do you find?
$$\displaystyle e * 0$$ ?

#### MarkFL

Staff member
$$\displaystyle e * 0$$ ?
Correct, and any constant times zero is zero.

You would get the same result if you assume $e^{x^0}=e$, since the derivative of a constant is zero, but I wanted you to be careful, as I felt the intent of the author of the problem was to not make that assumption.

#### wishmaster

##### Active member
Correct, and any constant times zero is zero.

You would get the same result if you assume $e^{x^0}=e$, since the derivative of a constant is zero, but I wanted you to be careful, as I felt the intent of the author of the problem was to not make that assumption.
Just to mention, $e$ represents natural number.

My solution for the first one is $$\displaystyle 3x^2-10x+4$$

Can you help me with second one?