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

1 Answer
Jan 22, 2018

Answer:

#d/dx x^(lnx) = (2lnx) x^(lnx-1)#

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

Explanation:

Write the function as:

#x^(lnx) = (e^lnx)^(lnx) = e^(ln^2x)#

then using the chain rule:

#d/dx x^(lnx) = e^(ln^2x) d/dx (ln^2x) = x^(lnx)(2lnx)/x#

and using the product rule:

#(d^2)/dx^2 x^(lnx) = (2lnx)/x d/dx x^(lnx) + x^(lnx) d/dx (2lnx)/x#

#(d^2)/dx^2 x^(lnx) = (4ln^2x)/x^2 x^(lnx) + 2x^(lnx) (1-lnx)/x^2#

#(d^2)/dx^2 x^(lnx) = x^(lnx) (4ln^2x -2lnx + 2)/x^2#

Note now that:

#x^(lnx)/x = x^(lnx)*x^-1 = x^(lnx-1)#

so we can simplify as:

#d/dx x^(lnx) = (2lnx) x^(lnx-1)#

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