Question

How do you differentiate $y = \ln (3x)$?

Answer 1

Here there is no direct formula for calculating the derivative of the given term; we will use the chain rule to find the derivative of the equation. On doing some simplification we get the required answer.

Complete step-by-step solution:
We have the given equation as:
$\Rightarrow y = \ln (3x)$
Now since there is no direct formula for calculating the derivative of the given expression, we will use the chain rule by writing the term as:
$\Rightarrow y' = \dfrac{d}{{dx}}\ln (3x)$
In this question we will consider $g(x) = 3x$
Now we know that the formula for the chain rule is: $F'(x) = f'(g(x))g'(x)$
Now we know that $\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}$, therefore on using the chain rule, we get:
$\Rightarrow y' = \dfrac{1}{{3x}}\dfrac{d}{{dx}}(3x)$.
Now we know that a constant is not the part of the derivative, therefore on taking $3$ out we can write the equation as:
$\Rightarrow y' = \dfrac{{1 \times 3}}{{3x}}\dfrac{d}{{dx}}(x)$
Now on simplifying the equation, we get:
$\Rightarrow y' = \dfrac{1}{x}\dfrac{d}{{dx}}(x)$
Now we know that $\dfrac{d}{{dx}}x = 1$, therefore on differentiating, we get:
$\Rightarrow y' = \dfrac{1}{x} \times 1$
On simplifying, we get:

$\Rightarrow y' = \dfrac{1}{x}$, which is the required solution.

Note: All the basic derivative formulas should be remembered to solve these types of sums, also whenever there is a constant value in multiplication in a derivative, it should be taken out of the derivative.
The inverse of the derivative is the integration and vice versa. If the derivative of a term $a$ is $b$, then the integration of the term $b$ will be $a$.
The term $\ln 3x$ represents the natural log of the term, the natural log has a base of $e$. The other most commonly used log is log to the base $10$. It is written as ${\log _{10}}x$. The base represents the number to which the log value should be raised to get the original value.