Question

What is the derivative of $\dfrac{1}{x\ln x}$ ?

Answer 1

Hint: We can find the derivative of the given function by using the product rule and quotient rule of differentiation. First, we use quotient rule which states that,
$\dfrac{d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}{dx}=\dfrac{g\left( x \right)\dfrac{d\left( f\left( x \right) \right)}{dx}-f\left( x \right)\dfrac{d\left( g\left( x \right) \right)}{dx}}{{{\left( g\left( x \right) \right)}^{2}}}$
 And then finally we can use the product rule to find the solution.
$\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\left( x \right)\dfrac{d\left( g\left( x \right) \right)}{dx}+g\left( x \right)\dfrac{d\left( f\left( x \right) \right)}{dx}$

Complete step by step solution:
In the question we are given the function, $\dfrac{1}{x\ln x}$ . We have been asked to find its derivative, that is, we need to find, $\dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}$ . First, we can use the quotient rule of differentiation to find the derivative of the given function. The quotient rule states that, for two function f and g, $\left( g\left( x \right)\ne 0 \right)$
$\dfrac{d\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)}{dx}=\dfrac{g\left( x \right)\dfrac{d\left( f\left( x \right) \right)}{dx}-f\left( x \right)\dfrac{d\left( g\left( x \right) \right)}{dx}}{{{\left( g\left( x \right) \right)}^{2}}}$
We can clearly see that in our case, $f\left( x \right)$ is 1 and $g\left( x \right)$ is $x\ln x$ . Therefore, by applying the rule, we get,
\[\dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=\dfrac{x\ln x\dfrac{d\left( 1 \right)}{dx}-1\dfrac{d\left( x\ln x \right)}{dx}}{{{\left( x\ln x \right)}^{2}}}\]
We know that the derivative of a constant is 0, hence, we end up with,
$\dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=\dfrac{-\dfrac{d\left( x\ln x \right)}{dx}}{{{\left( x\ln x \right)}^{2}}}$
Finally, we can use the product rule of differentiation to find the derivative in the numerator. The product rule of differentiation states that, for two function f and g,
$\dfrac{d\left( f\left( x \right)g\left( x \right) \right)}{dx}=f\left( x \right)\dfrac{d\left( g\left( x \right) \right)}{dx}+g\left( x \right)\dfrac{d\left( f\left( x \right) \right)}{dx}$
In our case, we can clearly see that $f\left( x \right)$ is x and $g\left( x \right)$ is $\ln x$ . Also, we know that the derivative of $\ln x$ is $\dfrac{1}{x}$ . Hence, we get,
$\begin{align}
  & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=\dfrac{-\left( x\dfrac{d\left( \ln x \right)}{dx}+\ln x\dfrac{d\left( x \right)}{dx} \right)}{{{\left( x\ln x \right)}^{2}}} \\
 & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=\dfrac{-\left( x\dfrac{1}{x}+\ln x\left( 1 \right) \right)}{{{\left( x\ln x \right)}^{2}}} \\
 & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=-\dfrac{1+\ln x}{{{\left( x\ln x \right)}^{2}}} \\
\end{align}$
Therefore, the derivative of the given function is found to be equal to $-\dfrac{1+\ln x}{{{\left( x\ln x \right)}^{2}}}$

Note: Another direct way to solve this problem, is by assuming $x\ln x$ to be let’s say y. Then we can write,
$\begin{align}
  & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=\dfrac{d\left( \dfrac{1}{y} \right)}{dx} \\
 & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=-\dfrac{1}{{{y}^{2}}}\times \dfrac{d\left( y \right)}{dx} \\
 & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=-\dfrac{1}{{{y}^{2}}}\times \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx} \\
 & \dfrac{d\left( \dfrac{1}{x\ln x} \right)}{dx}=-\dfrac{1+\ln x}{{{\left( x\ln x \right)}^{2}}} \\
\end{align}$
Therefore, the derivative of the given function using the alternate method is found to be $-\dfrac{1+\ln x}{{{\left( x\ln x \right)}^{2}}}$