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???
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) $
$\displaystyle = x^{a(x)} \cdot (a'(x) \ln x + a(x) \cdot \frac 1 x)$
$\displaystyle = x^{a(x)} a'(x) \ln x + a(x) x^{a(x) - 1}$
Logarithmic differentiation works well here...
$\displaystyle \displaystyle \begin{align*} y &= x^{a} \\ \ln{(y)} &= \ln{\left( x^a \right)} \\ \ln{(y)} &= a\ln{(x)} \\ \frac{d}{dx} \left[ \ln{(y)} \right] &= \frac{d}{dx} \left[ a\ln{(x)} \right] \\ \frac{d}{dy}\left[ \ln{(y)} \right] \frac{dy}{dx} &= a\,\frac{d}{dx} \left[ \ln{(x)} \right] + \ln{(x)} \,\frac{d}{dx} \left( a \right) \\ \frac{1}{y} \, \frac{dy}{dx} &= a \left( \frac{1}{x} \right) + \ln{(x)}\,\frac{da}{dx} \\ \frac{dy}{dx} &= y \left[ \frac{a}{x} + \ln{(x)} \,\frac{da}{dx} \right] \\ \frac{dy}{dx} &= x^a \left[ \frac{a}{x} + \ln{(x)} \, \frac{da}{dx} \right] \end{align*}$
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:
$\displaystyle e^{\ln(x^a)} = e^{a \ln x}$