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

[theakdad]

I have to derivate this function:

\(\displaystyle x^3-5x^2+4x-e^x\) 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:

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

Thank you all for the help!

 

[MarkFL]

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}.

 

[theakdad]

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 :D

 

[MarkFL]

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

 

[theakdad]

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

Its $e$¹

 

[MarkFL]

Its $e$¹

Is this true for $x=0$?

 

[theakdad]

Is this true for $x=0$?

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

 

[MarkFL]

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 $0^0=1$ ?

 

[theakdad]

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]

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?

 

[theakdad]

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]

\(\displaystyle e * 0 \) ?

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.

 

[theakdad]

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?