Question

What is the derivative of $f\left( x \right) = \ln {x^3}$?

Answer 1

Here, we have to find the derivative of $f\left( x \right) = \ln {x^3}$. In this equation, there are two functions $f\left( x \right) = \ln {x^3}$ and $g\left( x \right) = {x^3}$. So, to find the derivative of $f\left( x \right) = \ln {x^3}$ we will be using the chain rule. The chain rule is
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
So, we will be differentiating $\ln {x^3}$ and ${x^3}$ both.

Complete step by step solution:
In this question, we have to find the derivative of the function
$f\left( x \right) = \ln {x^3}$- - - - - - - - (1)
Now, there are two functions in the above equation.
So, when we are given more than one function in any equation, we need to use the chain rule to differentiate the equation.
According to the chain rule,
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Where, $f\left( x \right)$ and $g\left( x \right)$ are two functions.
According to our question,
$f\left( x \right) = \ln {x^3}$ and $g\left( x \right) = {x^3}$
So, we have to differentiate $\ln {x^3}$ and ${x^3}$ both.
Now, we know the derivative of $\ln x = \dfrac{1}{x}$ and derivative of ${x^3}$ is $3{x^2}$.
Therefore, equation (1) becomes
$\Rightarrow f\left( x \right) = \dfrac{d}{{dx}}\ln {x^3}$
$\Rightarrow f\left( x \right) = \dfrac{d}{{dx}}\ln {x^3} \cdot \dfrac{d}{{dx}}{x^3} \\\ \Rightarrow f\left( x \right) = \dfrac{1}{{{x^3}}} \times 3{x^2} \\\ \Rightarrow f\left( x \right) = \dfrac{3}{x} \\\$
Hence, the derivative of $f\left( x \right) = \ln {x^3}$ is $\dfrac{3}{x}$.

Note:
We learned the chain rule in these question, but there are two more important rules that we should learn. These two rules are

1. Product rule
2. Quotient rule
   Product rule: The product rule tells us how to differentiate the product of two different functions.
   $\Rightarrow$ $\dfrac{d}{{dx}}uv = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$
   For Example: $\dfrac{d}{{dx}}2x\cos x = 2x\dfrac{d}{{dx}}\cos x + \cos x\dfrac{d}{{dx}}2x = - 2x\sin x + 2\cos x$
   Quotient rule: Quotient rule is used for differentiating problems when one function is divided by another function.
   $\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{d}{{dx}}u - u\dfrac{d}{{dx}}v}}{{{v^2}}}$
   For example: $\dfrac{d}{{dx}}\left( {\dfrac{{2x}}{{\cos x}}} \right) = \dfrac{{\cos x\dfrac{d}{{dx}}2x - 2x\dfrac{d}{{dx}}\cos x}}{{{{\cos }^2}x}} = \dfrac{{2\cos x + 2x\sin x}}{{{{\cos }^2}x}}$