# Why the formula for derivatives

1. Mar 16, 2005

### 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?

2. Mar 16, 2005

### 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.

3. Mar 16, 2005

### Oggy

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

4. Mar 16, 2005

### 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

$$\int x^{z} \ dx = x^{z}\cdot x-\int (x^{z})'\cdot x \ dx =x^{z+1}-z\int x^{z} \ dx\Rightarrow \int x^{z} \ dx =\frac{1}{z+1} x^{z+1} +C ,z\neq -1$$

q.e.d.

Daniel.

Last edited: Mar 16, 2005
5. Mar 16, 2005

### Sterj

Why is

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

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

thanks

6. Mar 16, 2005

### 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.

7. Mar 17, 2005

### 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.