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?