How do you find the first and second derivative of #(lnx)^(x)#?

Question

Elijah Abram · Accepted Answer

Let:

#y=(lnx)^x#

The easiest way to differentiate this is to take the natural logarithm of both sides.

#ln(y)=ln((lnx)^x)#

Simplify using the logarithm rule:

#ln(y)=xln(lnx)#

Differentiate both sides. The left-hand side will need chain rule, similar to implicit differentiation, and the right-hand side will need chain rule and product rule.

#1/y*dy/dx=d/dx(x)*ln(lnx)+x*d/dxln(lnx)#

#1/y*dy/dx=ln(lnx)+x*1/lnx*d/dxlnx#

#1/y*dy/dx=ln(lnx)+x*1/lnx*1/x#

#1/y*dy/dx=ln(lnx)+1/lnx#

#1/y*dy/dx=(lnx*ln(lnx)+1)/lnx#

#dy/dx=(y(lnx*ln(lnx)+1))/lnx#

#dy/dx=((lnx)^x(lnx*ln(lnx)+1))/lnx#

#dy/dx=(lnx)^(x-1)(lnx*ln(lnx)+1)#

The steps to find the second derivative are far too lengthy, and the work is rather pointless, but if you wish try to find it, the second derivative is:

#(d^2y)/(dx^2)=((lnx)^(x-2)(lnx(xln(lnx)(lnx*ln(lnx)+2)+1)+x-1))/x#

Charles Arocha · Answer

undefined To find the first derivative of $ (\ln(x))^x $, we use the chain rule:

$$ \frac{d}{dx}\left((\ln(x))^x\right) = x(\ln(x))^{x-1}\left(\frac{1}{x}\right) + (\ln(x))^x\left(\frac{d}{dx}(x\ln(x))\right) $$

To find the second derivative, we differentiate the first derivative with respect to $ x $ again:

$$ \frac{d^2}{dx^2}\left((\ln(x))^x\right) = \frac{d}{dx}\left( x(\ln(x))^{x-1}\left(\frac{1}{x}\right) + (\ln(x))^x\left(\frac{d}{dx}(x\ln(x))\right) \right) $$