How do you differentiate #f(x)=ln(x)^x-lnx/x^x#?

1 Answer
Nov 9, 2015

#y' = {(1 - x^-x)/x} + ln(x){(1+ln(x))(x^-x)}#

Explanation:

Start with assuming #f(x) = y#.

Therefore,

#y = ln(x) - ln(x)/x^x#

#:. y = ln(x)[1 - 1/x^x]#

#:. y = ln(x)[1 - x^-x]#

Now, differentiating wrt x,

#y' = d/dx(ln(x){1 - x^-x})#

#:. y' = (1 - x^-x)d/dxln(x) + ln(x)d/dx(1 - x^-x)# ...(1)

(Using the Chain Rule of Differentation)

The first term is simple to handle, and it reduces to #(1 - x^-x)/x# but the second term is tricky, so we'll evaluate it separately.

Let #t = x^-x#

Taking the natural log on both sides of the equation,

# ln(t) = -xln(x) #

Differentiating both sides wrt x,

#1/tdt/dx = -(1 + ln(x))#

#:. dt/dx = -(1 + ln(x)).t#

Replacing value of t,

#d/dx(x^-x) = -(1 + ln(x))(x^-x)#

Extrapolating further and using the value #1 - x^-x#

#d/dx(1 - x^-x) = (1+ln(x))(x^-x)#

Using the above expression in (1),

#y' = {(1 - x^-x)/x} + ln(x){(1+ln(x))(x^-x)}#

This can be further simplified to suit the provided answer. But this, I believe, is in a simple form by itself. The trick here is to differentiate the #(1-x^-x)# separately, as shown, by equating it to the natural logarithm of any variable #t# and then simplifying the process of differentiation.