MHB What is the derivative of ln(x + √(1 + x^-25))?

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I have to derivate this function:

$$x^3-5x^2+4x-e^x$$ where $x$ on e is actualy $x^0$,but i don't 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:

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

Thank you all for the help!
 
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Just so we're clear, you are given:

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

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}.
 
MarkFL said:
Just so we're clear, you are given:

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

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 :D
 
Can you simplify $e^{x^0}$?
 
MarkFL said:
Can you simplify $e^{x^0}$?

Its $e$1
 
wishmaster said:
Its $e$1

Is this true for $x=0$?
 
MarkFL said:
Is this true for $x=0$?

So then $x$ is zero,and $e$ is also $0$ ?
 
wishmaster said:
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:

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

when $x=0$ ?

In other words, can we state $0^0=1$ ?
 
MarkFL said:
No, $e$ is a transcendental constant (like $\pi$), but what I was getting at, can we say:

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

when $x=0$ ?

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

So then remains just $e$ when i derivate?
 
  • #10
wishmaster said:
So then remains just $e$ when i derivate?

No, apply the exponential and chain rules:

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

What do you find?
 
  • #11
MarkFL said:
No, apply the exponential and chain rules:

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

What do you find?

$$e * 0 $$ ?
 
  • #12
wishmaster said:
$$e * 0 $$ ?

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.
 
  • #13
MarkFL said:
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 $$3x^2-10x+4$$

Can you help me with second one?
 

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