Please, what woudl be the derivative of a function?

I foudn sth like:

$\displaystyle (f(x)=x^a$

$\displaystyle (f(x)'=x^a(a)'$

is that ok???

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- Feb 4th 2013, 06:46 AMBooderivative of a function
Please, what woudl be the derivative of a function?

I foudn sth like:

$\displaystyle (f(x)=x^a$

$\displaystyle (f(x)'=x^a(a)'$

is that ok??? - Feb 4th 2013, 06:53 AMPlatoRe: derivative of a function
- Feb 4th 2013, 08:43 AMBooRe: derivative of a function
Hello!

I ment, if a = function!

some tol dme it is different!!! - Feb 4th 2013, 10:01 AMILikeSerenaRe: derivative of a function
Yep, that is different.

$\displaystyle (x^{a(x)})' $

$\displaystyle = (e^{a(x) \ln x})' $

$\displaystyle = e^{a(x) \ln x} \cdot (a'(x) \ln x + a(x) \cdot \frac 1 x) $

$\displaystyle = x^{a(x)} \cdot (a'(x) \ln x + a(x) \cdot \frac 1 x)$

$\displaystyle = x^{a(x)} a'(x) \ln x + a(x) x^{a(x) - 1}$ - Feb 4th 2013, 11:32 PMBooRe: derivative of a function
Yes, that is exactly what i ment!!

Do we "add" lnx in that formula or is that a typo? Many thans! I ment, is the original

$\displaystyle x^a$or $\displaystyle x^{alnx}$ - Feb 4th 2013, 11:44 PMProve ItRe: derivative of a function
Logarithmic differentiation works well here...

$\displaystyle \displaystyle \begin{align*} y &= x^{a} \\ \ln{(y)} &= \ln{\left( x^a \right)} \\ \ln{(y)} &= a\ln{(x)} \\ \frac{d}{dx} \left[ \ln{(y)} \right] &= \frac{d}{dx} \left[ a\ln{(x)} \right] \\ \frac{d}{dy}\left[ \ln{(y)} \right] \frac{dy}{dx} &= a\,\frac{d}{dx} \left[ \ln{(x)} \right] + \ln{(x)} \,\frac{d}{dx} \left( a \right) \\ \frac{1}{y} \, \frac{dy}{dx} &= a \left( \frac{1}{x} \right) + \ln{(x)}\,\frac{da}{dx} \\ \frac{dy}{dx} &= y \left[ \frac{a}{x} + \ln{(x)} \,\frac{da}{dx} \right] \\ \frac{dy}{dx} &= x^a \left[ \frac{a}{x} + \ln{(x)} \, \frac{da}{dx} \right] \end{align*}$ - Feb 5th 2013, 12:16 AMILikeSerenaRe: derivative of a function
- Feb 5th 2013, 01:28 AMBooRe: derivative of a function
got it!!!

many many thanks!!! - Feb 5th 2013, 01:45 AMILikeSerenaRe: derivative of a function
It follows from the calculation rules for logarithm and exponentials.

Suppose $\displaystyle q=\ln p$, then:$\displaystyle p = e^q$ by definition.If we substitute q, we get:[1] $\displaystyle p = e^{\ln p}$

We also have the rule:[2] $\displaystyle \ln p^q = q \ln p$

So to make that first step more explicit.

Using [1] we have:

$\displaystyle x^a = e^{\ln(x^a)}$And using [2] we have:

$\displaystyle e^{\ln(x^a)} = e^{a \ln x}$