What is the derivative of #f(x) = 1/xlnx#?

1 Answer
Oct 31, 2015

#(1-ln(x))/x^2#

Explanation:

Use the product rule: you have that

#d/dx f(x)*g(x) = f'(x)*g(x) + f(x)*g'(x)#.

Let's write down the quantity we need:

  • #f(x)= 1/x#;
  • #g(x)=ln(x)#;
  • #f'(x)=-1/x^2#;
  • #g'(x) = 1/x#.

Plugging these functions into the product rule, you have

#-1/x^2 * ln(x) + 1/x * 1/x = -ln(x)/x^2+1/x^2 = (1-ln(x))/x^2#