Why Are These the Formulas for Derivatives and Logarithms?

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Discussion Overview

The discussion revolves around the formulas for derivatives and integrals, specifically focusing on the power rule for derivatives and the integral of 1/x leading to the natural logarithm. Participants explore the reasoning behind these formulas, their derivations, and the conditions under which they apply.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant states that the derivative of x^n is nx^(n-1) and the antiderivative is 1/(n+1)x^(n+1), questioning the reasoning behind these formulas.
  • Another participant explains that the derivative power rule can be reversed using the antiderivative power rule, providing an example with x^2 and noting complications for n=-1.
  • A third participant presents the derivative and integral formulas in a more formal mathematical notation, referencing the generalized binomial formula and the Gamma function.
  • One participant asks for the derivation of the integral of 1/x, expressing difficulty in finding it online.
  • A later reply discusses the Fundamental Theorem of Calculus (FTC) and shows that the derivative of ln(x) is 1/x, leading to the conclusion that the integral of 1/x is ln(x) + C.
  • Another participant mentions that the integral from 1 to x of 1/t dt is sometimes used as the definition of ln(x) and suggests using implicit differentiation as an alternative method to derive the relationship.

Areas of Agreement / Disagreement

Participants present various methods and reasoning for the formulas discussed, but there is no explicit consensus on a single derivation or explanation. Multiple approaches and interpretations are offered, indicating a lack of agreement on a definitive explanation.

Contextual Notes

Some participants note that certain cases, such as n=-1, require different methods, and the discussion includes references to advanced mathematical concepts like the Gamma function and implicit differentiation, which may not be universally understood.

Who May Find This Useful

This discussion may be useful for students and enthusiasts of calculus, particularly those interested in the derivation and understanding of derivative and integral formulas.

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The formula for finding a derivative for x^n is nx^(n-1) and the anti derivative is 1/(n+1) x^(n+1)
Why is this the formula?
 
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If you take the derivative of a function using the (derivative) power rule, then you can always reverse it using the antiderivative power rule, its mainly just a reverse of it.

Like if you were to take the derivative of [tex]x^2[/tex], then using the power rule it would be [tex]2x^1[/tex] or just 2x. If you were to find the antiderivative of that using [tex]\frac{x^n+1}{n+1}[/tex] (agh supposed to be x^(n+1) not (x^n)+1) then it would be [tex]x^2[/tex] again.

If you're asking how it works, the derivative power rule derives itself from the limit of [tex]\frac{f(x+h) - f(h)}{h}[/tex] as h approaches 0. The antiderivative is a little more complicated and doesn't work in all instances (like when n=-1) so some other methods like natural logarithms need to be used.
 
[tex]\frac{d}{dx}x^{n}=nx^{n-1} \Rightarrow \int{x^{n-1}dx}=\frac{x^{n}}{n}[/tex]
 
The general rule of

[tex](x^{z})'=z x^{z-1},z\in \mathbb{C}[/tex]

is proven using the definition & generalized binomial formula (the one with Gamma Euler/Pochhammer symbols).

Once u've proven the Leibniz rule & implicitely the part integration mechanism,u can use the latter to

[tex]\int x^{z} \ dx = x^{z}\cdot x-\int (x^{z})'\cdot x \ dx =x^{z+1}-z\int x^{z} \ dx\Rightarrow \int x^{z} \ dx =\frac{1}{z+1} x^{z+1} +C ,z\neq -1<br /> [/tex]

q.e.d.

Daniel.
 
Last edited:
Why is

integral(1/x dx)=ln(x) ?

I can't find the derivation of this in google, afgh.

thanks
 
I've proven the general case for a complex exp.other than "-1".For this singular case,i'm using the FTC which says

[tex]\int f(x) \ dx=F(x)+C \Rightarrow \frac{dF(x)}{dx}=f(x)[/tex]

Then i know that

[tex]\frac{d\ln x}{dx} =\frac{1}{x}[/tex]

Ergo

[tex]\int \frac{1}{x} \ dx=\ln x+C[/tex]

Daniel.
 
Sometimes
[tex]\int_1^x \frac{1}{t}dt = \ln(x)[/tex]
is used as the definition for [itex]\ln x[/itex].

Otherwise, you can use implicit differentiation:

[tex]y=\ln x \iff x = e^y \Rightarrow 1=e^y \frac{dy}{dx} \iff \frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}[/tex]

and apply the FTC.
 

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