# logarithms, confusing

• Feb 24th 2008, 08:18 AM
Stylis10
logarithms, confusing
Given that y = X^x , x>0, y>0, by taking logs, show that

dy/dx = (X^x)(1+lnx)
• Feb 24th 2008, 08:29 AM
TheEmptySet
Quote:

Originally Posted by Stylis10
Given that y = X^x , x>0, y>0, by taking logs, show that

dy/dx = (X^x)(1+lnx)

Rewriting using logs
$lny=xlnx$

Taking the derivative
$\frac{1}{y}\frac{dy}{dx}=lnx+x\frac{1}{x}$

solving for y' gives

$\frac{dy}{dx}=y(lnx+1)$

sub in what y is equal to and vola!

$\frac{dy}{dx}=x^x(lnx+1)$
• Feb 24th 2008, 08:31 AM
earboth
Quote:

Originally Posted by Stylis10
Given that y = X^x , x>0, y>0, by taking logs, show that

dy/dx = (X^x)(1+lnx)

$y = x^x$ ............. Calculate the logarithms of both sides:

$\ln(y) = \ln(x^x) = x \cdot \ln(x)$ ............. Differentiate. Use the chainrule at the LHS and the product rule at the RHS:

$\frac1y \cdot \frac{dy}{dx} = x \cdot \frac1x + \ln(x) \cdot 1$ ............. Simplify:

$\frac{dy}{dx} = y \cdot (1 + \ln(x))$ Substitute y by the term of the function:

$\frac{dy}{dx} = x^x \cdot (1 + \ln(x))$