Question

What is the derivative of $\ln \left( 6x \right)$?

Answer 1

Hint: In this problem we need to calculate the derivative of the given function. In the given value we have logarithmic function. The value in the logarithmic function is $6x$ which is the product of the two variables $6$ and $x$. So, we will apply the logarithmic formula $\ln \left( ab \right)=\ln a+\ln b$. By applying this formula, we can write the given value as the sum of two values which are $\ln 6$ and $\ln x$. Now we will differentiate the obtained equation with respect to variable $x$. Here we will use the differentiation formula $\dfrac{d}{dx}\left( a+b \right)=\dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}\left( b \right)$. After applying this formula, we will simplify the obtained equation by using the differentiation values $\dfrac{d}{dx}\left( \text{constant} \right)=0$, $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$.

Complete step-by-step solution:
Given value is $\ln \left( 6x \right)$.
In the above value we can observe the product of the variables $6$ and $x$ are written in logarithmic function. By applying the logarithmic formula $\ln \left( ab \right)=\ln a+\ln b$ we can write the given value as
$\ln \left( 6x \right)=\ln 6+\ln x$
We know that the value of $\ln 6$ is $1.79$. Substituting this value in the above equation, then we will get
$\ln \left( 6x \right)=1.79+\ln x$
Differentiating the above equation with respect to the variable $x$, then we will have
$\dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=\dfrac{d}{dx}\left( 1.79+\ln x \right)$
Applying the differentiation formula $\dfrac{d}{dx}\left( a+b \right)=\dfrac{d}{dx}\left( a \right)+\dfrac{d}{dx}\left( b \right)$ in the above equation, then the above equation is modified as
$\dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=\dfrac{d}{dx}\left( 1.79 \right)+\dfrac{d}{dx}\left( \ln x \right)$
We know that the value $1.79$ is a constant. So, applying the differentiation formulas $\dfrac{d}{dx}\left( \text{constant} \right)=0$, $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ in the above equation, then we will get
$\begin{align}
  & \dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=0+\dfrac{1}{x} \\
 & \therefore \dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=\dfrac{1}{x} \\
\end{align}$
Hence the derivative of the value $\ln \left( 6x \right)$ is $\dfrac{1}{x}$.

Note: We can also directly solve this problem by using the differentiation formulas $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( \ln \left( ax \right) \right)=\dfrac{1}{ax}$, $\dfrac{d}{dx}\left( ax \right)=a$. Applying all these formulas to find the derivative of the given value $\ln \left( 6x \right)$, then we will get
$\begin{align}
  & \dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=\dfrac{d}{dx}\left( \ln \left( 6x \right) \right)\times \dfrac{d}{dx}\left( 6x \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=\dfrac{1}{6x}\times 6 \\
 & \Rightarrow \dfrac{d}{dx}\left( \ln \left( 6x \right) \right)=\dfrac{1}{x} \\
\end{align}$
From both the methods we got the same result.