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

Hint: 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.