How do you differentiate \[\ln \left( {3x} \right)\] ?

Hint: In this question, we have a composite function which is differentiable. To differentiate the composite function, we used the chain rule. And chain rule is given below. If the \[f\left( x \right)\] is a composite function. Then,
\[f\left( x \right) = \left( {g.h} \right)\left( x \right) = g\left[ {h\left( x \right)} \right]\]
Then, according to chain rule this function is different from below.
\[{f^{'}}\left( x \right) = {g^{'}}\left[ {h\left( x \right)} \right].{h^{'}}\left( x \right)\]

Complete step by step answer:
In this question, we used the word composite function. First we know about composite function. The composite function is defined as the function, which value is found from two given functions and apply one function to an independent variable and apply the second function to the result.
Then we come to the chain rule. The chain rule is defined as the technique for finding the derivative of a composite function.
Let’s take an example. If \[f\left( x \right)\]is a composite function, \[g\]and \[h\]is other two function then
Composite function \[f\left( x \right)\]is defined as.
\[f\left( x \right) = \left( {g.h} \right)\left( x \right) = g\left[ {h\left( x \right)} \right]\]
Now, we apply the chain rule to find the differentiation of that function.
Then,
\[ \Rightarrow {f^{'}}\left( x \right) = {g^{'}}\left[ {h\left( x \right)} \right]{h^{'}}\left( x \right)\]
Now we come to the question. In the question the function \[\ln \left( {3x} \right)\] is given.
Let's assume that the\[y\]is the function of \[x\]and equal to the function of \[\ln \left( {3x} \right)\].
Then, it is written as below.
\[ \Rightarrow y = \ln \left( {3x} \right)\]
Now we assume that the other function \[g\left( x \right)\]is equal to the\[3x\].
Then, it is written as below.
\[ \Rightarrow g\left( x \right) = 3x\]
Hence,
\[ \Rightarrow y = y\left( {g\left( x \right)} \right)\]
Where,
\[ \Rightarrow f\left( x \right) = \ln x\]
Then according to chain rule, the differentiation \[\left( {\dfrac{{dy}}{{dx}}} \right)\] is below that.
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dg}} \times \dfrac{{dg}}{{dx}}\]
Hence,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\ln \left( {3x} \right) = \dfrac{1}{{3x}} \times 3\]
Then,
\[\therefore \dfrac{{dy}}{{dx}} = \dfrac{1}{x}\]

Therefore, the differentiation of the function \[\ln \left( {3x} \right)\] is \[\dfrac{1}{x}\].

Note:
If you have a composite function and you want to differentiate this function. Then you used the chain rule to differentiate that type function. According to chain rule, it exists for differentiating a function of another function.