Answer
385.5k+ views
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 $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$
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 $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)