Question

Assertion (A): The $n^{th}$ derivative of $\log (3x+4)$ is $\frac{3^n(-1{)}^{n-1}(n-1)!}{(3x+4{)}^n}$
Reason (R): lf $y=\log |ax+b|$ then $y_n=\frac{(-1{)}^{n-1}(n-1)!a^n}{(ax+b{)}^n}$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Answer 1

Answer: (A)

Both A and R are true and R is the correct explanation of A

$y=\log |ax+b|$ $\Rightarrow$$y^{\prime }=\frac{a}{ax+b}$

$y^{\prime \prime }=-\frac{a^2}{{(ax+b)}^2}$ , $y^{\prime \prime \prime }=\frac{2a^3}{{(ax+b)}^3}$

$y^4=\frac{-3!a^4}{{(ax+b)}^4}$ , $y^5=\frac{4!a^5}{{(ax+b)}^5}$

So $y^n=\frac{{(-1)}^{n-1}{a}^n(n-1)!}{{(ax+b)}^n}$

So through this our assertion is also correct.