Why Are These the Formulas for Derivatives and Logarithms?

[blah]

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?

 

[motai]

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

If you're asking how it works, the derivative power rule derives itself from the limit of \frac{f(x+h) - f(h)}{h} 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.

 

[Oggy]

\frac{d}{dx}x^{n}=nx^{n-1} \Rightarrow \int{x^{n-1}dx}=\frac{x^{n}}{n}

 

[dextercioby]

The general rule of

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

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

\int x^{z} \ dx = x^{z}\cdot x-\int (x^{z})&#039;\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 /> <br />

q.e.d.

Daniel.

 

[Sterj]

Why is

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

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

thanks

 

[dextercioby]

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

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

Then i know that

\frac{d\ln x}{dx} =\frac{1}{x}

Ergo

\int \frac{1}{x} \ dx=\ln x+C

Daniel.

 

[Galileo]

Sometimes
\int_1^x \frac{1}{t}dt = \ln(x)
is used as the definition for \ln x.

Otherwise, you can use implicit differentiation:

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}

and apply the FTC.