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# The $n^{th}$ derivative of $\log \left| {3x + 4} \right|$ is $\dfrac{{{3^n}{{( - 1)}^{n - 1}}(n - 1)!}}{{{{(3x + 4)}^n}}}$ Reason(R): IF $y = \log \left| {ax + b} \right|$ then ${y_n} = \dfrac{{{{( - 1)}^{n - 1}}(n - 1)!{a^n}}}{{{{(ax + b)}^n}}}$ A) Both A and R are true and R is the correct explanation of AB) Both A and R are true but R is not the correct explanation of A C) A is true but R is false D) A is false but R is true

Last updated date: 20th Jun 2024
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Hint: In this assertion and reason are given to find the nth derivative for that firstly we will understand the meaning of nth derivative. Taking the derivative of the function n number of times is known as nth derivative of function A general formula for all of the successive derivative exists this formula is called the nth derivative for this firstly we have to find the derivative of order $5$ from such results we can derive a general formula for higher order and that general formula is called the nth derivative

Formula used: Formula of derivative of $y = \log \left| {ax + b} \right|$ is $\dfrac{a}{{ax + b}}$
Basically it is based on $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$

Complete step-by-step solution:
step1: To find the derivative of $\log \left| {ax + b} \right|$ we will use the formula
$\dfrac{d}{{dx}}\log \left| {ax + b} \right| = \dfrac{a}{{ax + b}}$
So $y' = \dfrac{a}{{ax + b}}$ this is the first order derivative
Step2: Similarly we will find second order derivative
$\dfrac{d}{{dx}}\left( {\dfrac{a}{{ax + b}}} \right) =$
Using the quotient rule to solve this
Quotient rule is $\dfrac{d}{{dx}}\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{(g(x))}^2}}}$

$\dfrac{d}{{dx}}\left( {\dfrac{a}{{ax + b}}} \right) = \dfrac{{(ax + b)\dfrac{{d(a)}}{{dx}} - a\dfrac{d}{{dx}}(ax + b)}}{{{{(ax + b)}^2}}}$
On finding the derivate in numerator
$\Rightarrow \dfrac{{0 - {a^2}}}{{{{(ax + b)}^2}}} = \dfrac{{ - {a^2}}}{{{{(ax + b)}^2}}}$
To find the third derivative
$\left( {\dfrac{{dy}}{{dx}}} \right)\left( {\dfrac{{ - {a^2}}}{{{{(ax + b)}^2}}}} \right) = \dfrac{{{{(ax + b)}^2}\dfrac{{d( - {a^2})}}{{dx}} - ( - {a^2})\dfrac{d}{{dx}}{{(ax + b)}^2}}}{{{{(ax + b)}^4}}}$
On solving the equation we get
$= \dfrac{{0 + 2{a^2}(ax + b) \times a}}{{{{(ax + b)}^4}}}$
On dividing the equation by ${(ax + b)^4}$
$= \dfrac{{2{a^3}}}{{{{(ax + b)}^4}}}$
Similarly we get
$y'4 = \dfrac{{3!{a^4}}}{{{{(ax + b)}^4}}}$
And $y'5 = \dfrac{{4!{a^5}}}{{{{(ax + b)}^5}}}$
Step3: We can derive the nth formula will be
Derivatives of different order for $y = \log \left| {ax + b} \right|$is
$\Rightarrow {y_1} = \dfrac{a}{{ax + b}}$;
$\Rightarrow {y_2} = \dfrac{{( - 1)}}{{{{(ax + b)}^2}}}{a^2}$;
$\Rightarrow {y_3} = \dfrac{{( - 1)( - 2)}}{{{{(ax + b)}^4}}}{a^3}$;

$\Rightarrow {y_4} = \dfrac{{( - 1)( - 2)( - 3)}}{{{{(ax + b)}^4}}}{a^4}$;
$\Rightarrow {y_5} = \dfrac{{( - 1)( - 2)( - 3)( - 4)}}{{{{(ax + b)}^5}}}{a^5}$
$\Rightarrow$……. Similarly like this series will form of more higher order derivatives
By observing this we can derive the general formula for the ${n^{th}}$term
$\Rightarrow {y_n} = \dfrac{{( - 1)( - 2)( - 3)....( - n + 1)}}{{{{(ax + b)}^n}}}{a^n}$
$\Rightarrow {y_n} = \dfrac{{{{( - 1)}^{n - 1}}1.2.3...(n - 1)}}{{{{(ax + b)}^n}}}{a^n}$
On further solving the equation and making it in the general form:
${y_n} = \dfrac{{{{( - 1)}^{n - 1}}(n - 1)!}}{{{{(ax + b)}^n}}}{a^n}$
Hence the general formula for the ${n^{th}}$ term will be:
${y^n} = \dfrac{{{{( - 1)}^{n - 1}}{a^n}(n - 1)!}}{{{{(ax + b)}^n}}}$
On putting the value of variables from $\log \left| {3x + 4} \right|$ here $a = 3$ and $b = 4$
Then on substituting the values we will get the nth derivative for this will be: $\dfrac{{{3^n}{{( - 1)}^{n - 1}}(n - 1)!}}{{(3a + 4)}}$
In this reason and assertion both are correct

Option (A) is correct.

Note: In such problems students mainly do mistakes in finding the derivative they usually find the first order derivative right and they do some calculation error in finding 4th and 5th order derivative sometimes they apply the wrong quotient rule or they are not aware of it so just learn this formula
Commit to memory:
Formula of derivative of $y = \log \left| {ax + b} \right|$ is $\dfrac{a}{{ax + b}}$
Basically it is based on $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$