Question

# Find the derivatives of the functions. f(x)=\frac{1}{x\ln x}

Derivatives
Find the derivatives of the functions. $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}{\ln{{x}}}}}}$$

2021-05-18
The function we want to differentiate is
$$\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{x}{\ln{{x}}}}}}$$
Apply the quotient rule first to get
$$\displaystyle{f}'{\left({x}\right)}={\frac{{{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({1}\right)}{x}{\ln{{x}}}-{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}{\ln{{x}}}\right)}}}{{{\left({x}{\ln{{x}}}\right)}^{{{2}}}}}}$$
Note that the first term in the numerator is zero because
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({1}\right)}={0}$$
The second term can be computed using the product rule and the derivative of natural logarithm rule:
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\ln{{x}}}\right)}={\frac{{{1}}}{{{x}}}}$$
So we get
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}{\ln{{x}}}\right)}={\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({x}\right)}{\ln{{x}}}+{x}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({\ln{{x}}}\right)}$$
$$\displaystyle={\ln{{x}}}+{x}{\frac{{{1}}}{{{x}}}}$$
$$\displaystyle={\ln{{x}}}+{1}$$
Substituting back, we get
$$\displaystyle{f}'{\left({x}\right)}={\frac{{-{\left({\ln{{x}}}+{1}\right)}}}{{{\left({x}{\ln{{x}}}\right)}^{{{2}}}}}}$$