How do you differentiate $\ln {x^{\dfrac{1}{3}}}?$

Hint:Use chain rule to in order to solve this question, consider ${x^{\dfrac{1}{3}}}$ to be $u$ and then apply chain rule as follows $\dfrac{{df(u)}}{{du}} \times \dfrac{{du}}{{dx}}$ , where $f(u)$ is the given function in which ${x^{\dfrac{1}{3}}}$ is replaced by $u$
Also derivative of logarithm function is given as $\dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x}$

Complete step by step solution:
 In this type of question in which composite functions lie, we use chain rule in order to derive the composite function. If we have $y = f(u)\;{\text{and}}\;u = g(x)$ then the derivative of will be given as follows
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Now coming to the question, we have to find the derivative of $\ln {x^{\dfrac{1}{3}}}$,
In order to derive $\ln {x^{\dfrac{1}{3}}}$ we will consider ${x^{\dfrac{1}{3}}} = u$
Now we have
$
y = \ln {x^{\dfrac{1}{3}}}\;{\text{and}}\;u = {x^{\dfrac{1}{3}}} \\
\Rightarrow y = \ln u\;{\text{and}}\;u = {x^{\dfrac{1}{3}}} \\
$
Differentiating $y$ with respect to $x$ we will get
$\dfrac{{dy}}{{dx}} = \dfrac{{d\ln u}}{{dx}}$
And by chain rule we can write this as
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\ln u}}{{du}} \times \dfrac{{du}}{{dx}}$
Using the derivation $\dfrac{{d\ln x}}{{dx}} = \dfrac{1}{x}$ , we will get
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{d\ln u}}{{du}} \times \dfrac{{du}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{u} \times \dfrac{{du}}{{dx}} \\
$
Now putting the value of $u = {x^{\dfrac{1}{3}}}$ in the derivation and further derivate it with respect to $x$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^{\dfrac{1}{3}}}}} \times
\dfrac{{d{x^{\dfrac{1}{3}}}}}{{dx}}$
According to power rule $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
Using the power rule to find $\dfrac{{d{x^{\dfrac{1}{3}}}}}{{dx}}$, we will get
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^{\dfrac{1}{3}}}}} \times
\dfrac{{d{x^{\dfrac{1}{3}}}}}{{dx}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^{\dfrac{1}{3}}}}} \times \dfrac{1}{3} \times
{x^{\dfrac{1}{3} - 1}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3} \times \dfrac{1}{{{x^{\dfrac{1}{3}}}}} \times
{x^{\dfrac{{ - 2}}{3}}} \\
$
Now writing ${x^{\dfrac{{ - 2}}{3}}} = \dfrac{1}{{{x^{\dfrac{2}{3}}}}}$
$
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3} \times \dfrac{1}{{{x^{\dfrac{1}{3}}}}} \times
\dfrac{1}{{{x^{\dfrac{2}{3}}}}} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3} \times \dfrac{1}{{{x^{\dfrac{1}{3} + \dfrac{2}{3}}}}}
\\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{3} \times \dfrac{1}{x} \\
\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{3x}} \\
$
That is the required derivative of $\ln {x^{\dfrac{1}{3}}}$ is equals to $\dfrac{1}{{3x}}$

Note: Chain rule has one more application if functions $f(x)\;{\text{and}}\;g(x)$ are separately differentiable then the function $h(x) = f \circ g(x)$ will be differentiated as $h'(x) = f'(g(x)) \times g'(x)$
We can solve this problem directly without using the chain rule, by using the property of logarithm function.
If the argument of a log has some digit in its power then we can rewrite the log function as the exponent times the logarithm of the base. Mathematically it can be expressed as follows
$\ln {x^n} = n \times \ln x$
This rule is also known as the power rule of logarithm function.