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(d^n)/(dx^n)(logx)= ((n-1)!)/(x^n) (b) ...

`(d^n)/(dx^n)(logx)=` `((n-1)!)/(x^n)` (b) `(n !)/(x^n)` `((n-2)!)/(x^n)` (d) `(-1)^(n-1)((n-1)!)/(x^n)`

A

`((n-1)!)/(x^(n))`

B

`(n!)/(x^(n))`

C

`((n-2)!)/(x^(n))`

D

`(-1)^(n-1)((n-1)!)/(x^(n))`

Text Solution

AI Generated Solution

To solve the problem of finding the \( n \)-th derivative of \( \log x \), we can follow these steps: ### Step 1: Define the function Let \( y = \log x \). ### Step 2: Find the first derivative The first derivative of \( y \) with respect to \( x \) is: \[ ...
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(d^n)/(dx^n)(logx)=? (a) ((n-1)!)/(x^n) (b) (n !)/(x^n) (c) ((n-2)!)/(x^n) (d) (-1)^(n-1)((n-1)!)/(x^n)

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Knowledge Check

  • If (n^(n))/(n!)=(nx)/((n-1)!), x =

    A
    `n^(n-2)`
    B
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    C
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    D
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