proofs for the derivatives of logax, lnx, a^x and e^x

**normalguy** · April 10th 2011, 05:50 AM

I need short, sweet, simple proofs for all the derivatives of $log {a} {x}$ , $ln{x}$ , $a^x$ and $e^x$ concurrently so as to avoid a "chicken or the egg" situation in which 1 proof relies on one of the other 3 results.

Note:
${d(\log{a}{x}})/{dx} = 1/{x\ln{a}}$
$d(\ln{x})/{dx} = 1/x$
$d(a^x)/dx = a^x\ln{a}$
$d(e^x)/dx = e^x$

Caution: Do not commit circular reasoning like what I have encountered.

Anyone who can do so is likely to benefit most of us who are learning derivatives.

"Proof is what sets mathematics apart from every other science." ----Prove It, MHF Contributor

Thank You.

**awkward** · April 10th 2011, 05:57 AM

Hi normalguy,

It all depends on how you define those functions.

So, what is your definition of $\ln x$ ? How about $e^x$ ?

**normalguy** · April 10th 2011, 06:15 AM

Originally Posted by awkward

Hi normalguy,

It all depends on how you define those functions.

So, what is your definition of $\ln x$ ? How about $e^x$ ?

????

**skeeter** · April 10th 2011, 06:27 AM

Originally Posted by normalguy

????

definition of the transcendental number $e$ ...

$\displaystyle e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$

or

$\displaystyle e = \lim_{x \to 0} \left(1 + x\right)^{\frac{1}{x}}$

definition of the natural log function ...

$\displaystyle \ln{x} = \int_1^x \frac{1}{t} \, dt$

you have to start with one or the other.

**Prove It** · April 10th 2011, 06:48 AM

Start by defining $\displaystyle e^x$ as the function that is its own derivative (for that IS the definition of $\displaystyle e^x$ ).

The other proofs follow with some algebraic manipulation.

**normalguy** · April 10th 2011, 07:16 AM

Originally Posted by skeeter

definition of the transcendental number $e$ ...

$\displaystyle e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$

or

$\displaystyle e = \lim_{x \to 0} \left(1 + x\right)^{\frac{1}{x}}$

definition of the natural log function ...

$\displaystyle \ln{x} = \int_1^x \frac{1}{t} \, dt$

you have to start with one or the other.

can you explain further on this? why is e=(1+y)^y

**normalguy** · April 10th 2011, 07:18 AM

so is the proof of the derivative of logax dependent on the result of the derivative of lnx or vice versa?

**skeeter** · April 10th 2011, 07:39 AM

Originally Posted by normalguy

so is the proof of the derivative of logax dependent on the result of the derivative of lnx or vice versa?

if you accept the fact that $\dfrac{d}{dx} (\ln{x}) = \dfrac{1}{x}$ , then ...

$\dfrac{d}{dx} (\ln{ax}) = \dfrac{d}{dx} (\ln{a} + \ln{x}) = 0 + \dfrac{1}{x} = \dfrac{1}{x}$

Originally Posted by normalguy

can you explain further on this? why is e=(1+y)^y

that equation is not what I posted ... if you want what you call "a non-circular proof", then you need the necessary background information to understand the proof.

e (mathematical constant) - Wikipedia, the free encyclopedia

**Prove It** · April 10th 2011, 08:38 AM

I expect the OP meant $\displaystyle \log_a{x}$ , not $\displaystyle \log{ax}$ .

Notice that any logarithm can be changed to the natural logarithm...

$\displaystyle y = \log_a{x}$

$\displaystyle a^y = x$

$\displaystyle \ln{\left(a^y\right)} = \ln{x}$

$\displaystyle y\ln{a} = \ln{x}$

$\displaystyle y = \frac{\ln{x}}{\ln{a}}$ .

So what is the derivative of $\displaystyle \log_a{x}$ ?

**normalguy** · April 11th 2011, 06:43 AM

Originally Posted by Prove It

I expect the OP meant $\displaystyle \log_a{x}$ , not $\displaystyle \log{ax}$ .

Notice that any logarithm can be changed to the natural logarithm...

$\displaystyle y = \log_a{x}$

$\displaystyle a^y = x$

$\displaystyle \ln{\left(a^y\right)} = \ln{x}$

$\displaystyle y\ln{a} = \ln{x}$

$\displaystyle y = \frac{\ln{x}}{\ln{a}}$ .

So what is the derivative of $\displaystyle \log_a{x}$ ?

I got it. 1/xlna

so you are using d/dx(lnx)=1/x to proof the derivative of $\displaystyle \log_a{x}$

But I am curious to know which was proven first? derivative of $\displaystyle \log_a{x}$ or $ln{x}$ ?

My textbook says in the special case where a=e for $\displaystyle y = \log_a{x}$ , the derivative is 1/xlne=1/x. Hence I assume derivative of $log_a{x}$ should be proven first!

But your working seems to suggest otherwise.

**normalguy** · April 11th 2011, 07:03 AM

meanwhile, some1 explain this thingy $\displaystyle e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ which apparently seems to be the basis of all the proofs, right?

**Plato** · April 11th 2011, 07:05 AM

Originally Posted by normalguy

I got it. 1/xlna.

That is correct is read correctly.
You really do need to learn to use symbols.
Why not learn to post in symbols? You can use LaTeX tags
[tex]y=\log_a(x)~\Rightarrow y'=\dfrac{1}{x\ln(a)}[/tex] gives $y=\log_a(x)~\Rightarrow y'=\dfrac{1}{x\ln(a)}$

Not being able to read your post slows us down.

**Plato** · April 11th 2011, 07:28 AM

Originally Posted by normalguy

meanwhile, some1 explain this thingy $\displaystyle e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$ which apparently seems to be the basis of all the proofs, right?

There are several ways to approach this question. They start at different places and end up at that conclusion.
Defining $\mathbf{e}$ as the number such that $\int_1^e {\frac{{dx}}{x}} = 1$ ties $\mathbf{e}^x$ and $\ln(x)$ together.

The using the monotone property of integrals we can get $\displaystyle e^{\frac{n}{{n + 1}}} \leqslant \left( {1 + \frac{1}{n}} \right)^n \leqslant e$ .

The limit you asked about is the next step.

**Soroban** · April 11th 2011, 08:08 AM

Hello, normalguy!

Do you really want four separate proofs?
That is a lot of unnecessary work.
If you insist on four proofs, I suggest you find them yourself.

$\text{I need short/sweet/simple proofs for the derivatives of }\log_a x,\;\ln x,\;a^x$
$\text{ and }e^x\,\text{ concurrently so as to avoid a "chicken or the egg" situation}$
$\text{in which one proof relies on one of the other 3 results.}$

. . $\dfrac{d}{dx}(\log_ax) \:=\: \dfrac{1}{x\ln a}$

. . $\dfrac{d}{dx}(\ln x ) \:=\: \dfrac{1}{x}$

. . $\dfrac{d}{dx}(a^x) \:=\: a^x\ln{a}$

. . $\dfrac{d}{dx}(e^x) \:=\: e^x$
.

I'll prove the first one . . . The others will follow nicely.

We have: . $f(x) \;=\;\log_a(x)$

$f(x+h) - f(x) \;=\;\log_a(x+h) - \log_a(x) \;=\;\log_a\left(\dfrac{x+h}{x}\right)$

$\displaystyle \frac{f(x+h)-f(x)}{h} \;=\;\frac{1}{h}\log_a\left(\frac{x+h}{x}\right) \;=\; \log_a\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}$

. . . . . . . . . . . . . $\displaystyle =\;\log_a\left[\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}} \;=\;\frac{\ln\left[\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}}{\ln a}$

$\displaystyle f'(x) \;=\;\lim_{h\to0}\left\{ \frac{\ln\left[\left(1+\frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}}{\ln a}\right\} \;=\;\frac{\ln\overbrace{\left[\lim\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]}^{\text{This is }e}^{\frac{1}{x}}}{\ln a}$

. . . . . $\displaystyle =\;\frac{\ln(e^{\frac{1}{x}})}{\ln a} \;=\; \frac{\frac{1}{x}}{\ln a}$

$\displaystyle \text{Therefore: }\;\frac{dx}{dx}(\log_a x) \;=\;\frac{1}{x\ln a}$

**normalguy** · April 12th 2011, 04:06 AM

i manage to figure out the proofs of derivatives in the following order:
$1. ln(x)<br /> 2. \log_a x<br /> 3. e^x<br /> 4. a^x<br />$
am i right to say that the basis of all proofs is the definition of e?

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