Question

Find the value of $\dfrac{{{d^n}}}{{d{x^n}}}(\log x)$
$A)\dfrac{{(n - 1)!}}{{{x^n}}}$
$B)\dfrac{{n!}}{{{x^n}}}$
$C)\dfrac{{(n - 2)!}}{{{x^n}}}$
$D)\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{x^n}}}$

Answer 1

First, we shall analyze the given information so that we can able to solve the problem. Generally, in Mathematics, the derivative refers to the rate of change of a function with respect to a variable. Here, we are applying the power rule to find the required answer.
We often use the power rule to calculate the derivative of a variable raised to a power and the power rule is the most commonly used derivative rule.
Differentiation and integration are an inverse process where $\dfrac{d}{{dx}}({x^2}) = 2x$ and $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$
Formula to be used:
The formula that is applied in the power rule is as follows
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ (First differentiation)
$\dfrac{{{d^2}}}{{d{x^2}}}\left( {{x^n}} \right) = n(n - 1){x^{n - 2}}$ (Second differentiation)
$\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$

Complete step by step answer:
Here we are asked to find the derivative of ${n^{th}}$ the function $\log x$
Let us start with the first derivative of the given function, that is $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$ (the differentiation of $\log x$ with respect to x)
The second derivative of the given function is $\dfrac{{{d^2}}}{{d{x^2}}}\left( {\log x} \right) = \dfrac{d}{{dx}}(\dfrac{1}{x})$ $\Rightarrow \dfrac{{ - 1}}{{{x^2}}}$
This can be obtained $\dfrac{d}{{dx}}(\dfrac{1}{x}) = \dfrac{d}{{dx}}({x^{ - 1}})$ by using the first derivative formula, we get $\dfrac{d}{{dx}}({x^{ - 1}}) = - 1({x^{ - 2}})$ $\Rightarrow \dfrac{{ - 1}}{{{x^2}}}$
Hence the second derivative $\log x$ is $\Rightarrow \dfrac{{ - 1}}{{{x^2}}}$
Now to find the third derivative of the given function, $\dfrac{{{d^3}}}{{d{x^3}}}\left( {\log x} \right) = \dfrac{{{d^2}}}{{d{x^2}}}(\dfrac{1}{x}) = \dfrac{d}{{dx}}(\dfrac{{ - 1}}{{{x^2}}})$
Thus, we get $\dfrac{d}{{dx}}(\dfrac{{ - 1}}{{{x^2}}}) = \dfrac{d}{{dx}}( - {x^{ - 2}}) \Rightarrow 2{x^{ - 3}}$
We have derivate the log in three times, thus from this we can able to write the generalized form the function $\log x$
That is the derivate of ${n^{th}}$ to the function $\log x$ is $\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{x^n}}}$
Since apply $n = 1$ in the above equation, then we get $\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{x^n}}} = \dfrac{{{{( - 1)}^{1 - 1}}(1 - 1)!}}{{{x^1}}} \Rightarrow \dfrac{1}{x}$ which is the first derivative of $\log x$
Again apply $n = 2$ then we get, $\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{x^n}}} = \dfrac{{{{( - 1)}^{2 - 1}}(2 - 1)!}}{{{x^2}}} \Rightarrow \dfrac{{ - 1}}{{{x^2}}}$ which is the second derivative of the function $\log x$ , Therefore, the generalized ${n^{th}}$ derivative is $\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{x^n}}}$
Thus, $D)\dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{x^n}}}$ is correct.
Since for the options like $A)\dfrac{{(n - 1)!}}{{{x^n}}}$ apply $n = 2$ then we get $\dfrac{{(2 - 1)!}}{{{x^2}}} \Rightarrow \dfrac{1}{{{x^2}}}$ (which is not the second derivation of the function $\log x$ ), thus option A is incorrect.
$B)\dfrac{{n!}}{{{x^n}}}$ apply $n = 2$ then $\dfrac{{n!}}{{{x^n}}} = \dfrac{2}{{{x^2}}}$ thus option B is incorrect
$C)\dfrac{{(n - 2)!}}{{{x^n}}}$ to apply $n = 2$ then $\dfrac{{(n - 2)!}}{{{x^n}}} = \dfrac{1}{{{x^2}}}$ , option C is incorrect.

Note: The power rule is one of the derivative rules such that $x$ is a variable that is raised to a power $n$ , then the derivative of $x$ raised to the power is denoted by the formula, $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
After finding the ${n^{th}}$ derivative, just apply the value of n as two, to eliminate the other options.
Also, we often use the power rule to calculate the derivative of a variable raised to a power and the power rule is the most commonly used derivative rule.