# derivative of a function

• Feb 4th 2013, 06:46 AM
Boo
derivative 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 AM
Plato
Re: derivative of a function
Quote:

Originally Posted by Boo
Please, what woudl be the derivative of a function?
I foudn sth like:
$\displaystyle (f(x)=x^a$ find $\displaystyle (f(x)'=x^a(a)'$

If $\displaystyle a$ is a non-zero constant then

$\displaystyle \left(x^a\right)^{\prime}=a\left(x\right)^{a-1}$
• Feb 4th 2013, 08:43 AM
Boo
Re: derivative of a function
Hello!
I ment, if a = function!
some tol dme it is different!!!
• Feb 4th 2013, 10:01 AM
ILikeSerena
Re: 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 PM
Boo
Re: 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 PM
Prove It
Re: derivative of a function
Quote:

Originally Posted by Boo
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???

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 AM
ILikeSerena
Re: derivative of a function
Quote:

Originally Posted by Boo
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 $\displaystyle x^{alnx}$

There is no typo. lnx is in the formula, but it is not added but multiplied.
I guess I could also write it as:

$\displaystyle (x^a)' = x^a a' \ln x + ax^{a-1}$
• Feb 5th 2013, 01:28 AM
Boo
Re: derivative of a function
got it!!!
many many thanks!!!
• Feb 5th 2013, 01:45 AM
ILikeSerena
Re: derivative of a function
Quote:

Originally Posted by Boo
Yes, I see!
How did U get the first line, sorry, I really forgot...
$\displaystyle (x^a(x))'=(e^{a(x)lnx}$
Many thanks¨!!!

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:

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