# Logarithmic Diff.....

• January 27th 2007, 05:01 PM
Logarithmic Diff.....
Find the derivative of the function
$f(x) = x^{\sqrt{x}}$; $x>0$
Do i have to use logarithmic differentiation?
The answer according to the back of book is....
$x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)$
how?
• January 27th 2007, 05:13 PM
ThePerfectHacker
Quote:

Find the derivative of the function
$f(x) = x^{\sqrt{x}}$; $x>0$
Do i have to use logarithmic differentiation?
The answer according to the back of book is....
$x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)$
how?

You can write,
$x=e^{\ln x}$
Thus,
$x^{\sqrt{x}}=e^{\ln x\sqrt{x}}$
Find the derivative of the inside function,
$u=\ln x\sqrt{x}$
$u'=(\ln x)'\sqrt{x}+(\ln x)(\sqrt{x})'$
$u'=\frac{1}{x}\cdot \sqrt{x} + \ln x\cdot \frac{1}{2\sqrt{x}}$
Irrationalize the denominator,
$u'=\frac{1}{\sqrt{x}}+\frac{\ln x}{2\sqrt{x}}$
$u'=\frac{2+\ln x}{2\sqrt{x}}=\left( \frac{2+\ln x}{2} \right) \frac{1}{\sqrt{x}}=(1+\frac{1}{2}\ln x)x^{-1/2}$
Now times the outer function,
$(1+\frac{1}{2}\ln x)x^{-1/2}e^{\ln x\sqrt{x}}=(1+\frac{1}{2}\ln x)x^{-1/2}x^{\sqrt{x}}$
Laws of exponents,
$(1+\frac{1}{2}\ln x)x^{\sqrt{x} - \frac{1}{2}}$
• January 27th 2007, 10:08 PM
CaptainBlack
Quote:

Find the derivative of the function
$f(x) = x^{\sqrt{x}}$; $x>0$
Do i have to use logarithmic differentiation?
The answer according to the back of book is....
$x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)$
how?

$\ln(f(x))=\sqrt{x}\ln(x)$

so:

$\frac{1}{f(x)}\,\frac{df}{dx}=\frac{1}{2\sqrt{x}}\ ln(x)+\frac{\sqrt{x}}{x}$

so:

$
\frac{df}{dx}= \frac{x^{\sqrt{x}} \ln(x)}{2\sqrt{x}}+\frac{x^{\sqrt{x}}\sqrt{x}}{x}
$

which when all the powers of $x$ are collected together gives the required result:

$
\frac{df}{dx}= \frac{x^{\sqrt{x}-1/2} \ln(x)}{2}+x^{\sqrt{x}-1/2}
$

or:

$\frac{df}{dx}=x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)$
• January 28th 2007, 12:45 AM
ticbol
Quote:

Find the derivative of the function
$f(x) = x^{\sqrt{x}}$; $x>0$
Do i have to use logarithmic differentiation?
The answer according to the back of book is....
$x^{\sqrt{x} - \frac{1}{2}}(1 + \frac{1}{2} \ln x)$
how?

Here is one way.

f(x) = x^(sqrt(x))

Let y = f(x)
So,
y = x^(sqrt(x))
Take the ln of both sides,
ln(y) = sqrt(x) *ln(x)
Differentiate both sides with respect to x,
(1/y)y' = [sqrt(x) *1/x] +[ln(x) *(1/2)x^(-1/2)]
(1/y)y' = [1/sqrt(x)] +[1/(2sqrt(x)) *ln(x)]
Factor the RHS by 1/sqrt(x),
(1/y)y' = (1/sqrt(x))[1 +(1/2)ln(x)]
Multiply both sides by y,
And since 1/sqrt(x) = x^(-1/2),
y' = (x^sqrt(x))*(x^(-1/2))[1 +(1/2)ln(x)]
y' = [x^(sqrt(x) -(1/2)][1 +(1/2)ln(x)] -----answer at the back of the book.