You visited us 3 times! Enjoying our articles? Unlock Full Access!

Algebra of Derivatives

dn/dxn[logax+...

Question

$\frac{d^{n}}{d x^{n}} [l o g (a x + b)]$ is equal to:

\frac{(- 1)^{n} n! a^{n}}{(a x + b)^{n + 1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(- 1)^{n - 1} (n - 1)! a^{n}}{(a x + b)^{n + 1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(- 1)^{n + 1} (n + 1)! a^{n - 1}}{(a x + b)^{n + 1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{(- 1)^{n - 1} (n - 1)! a^{n}}{(a x + b)^{n}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $\frac{(- 1)^{n - 1} (n - 1)! a^{n}}{(a x + b)^{n}}$
Let $y = log (a x + b)$ , let $D = \frac{d}{d x}$
$\frac{d y}{d x} = D y = \frac{a}{a x + b}$
$D^{2} y = \frac{- a^{2}}{{(a x + b)}^{2}}, D^{3} y = \frac{2 a^{3}}{{(a x + b)}^{3}}$
$D^{4} y = \frac{- 3.2.1. a^{4}}{{(a x + b)}^{4}}$
$∴ D^{n} y = \frac{{(- 1)}^{n - 1} (n - 1)! a^{n}}{{(a x + b)}^{n}}$
$∴ \frac{d^{n}}{d x^{n}} (log (a x + b)) = \frac{{(- 1)}^{n - 1} (n - 1)! a^{n}}{{(a x + b)}^{n}}$

Suggest Corrections

Similar questions

n^{t h}

log (3 x + 4)

\frac{3^{n} (- 1)^{n - 1} (n - 1)!}{(3 x + 4)^{n}}

y = log | a x + b |

y_{n} = \frac{(- 1)^{n - 1} (n - 1)! a^{n}}{(a x + b)^{n}}

l o g a + l o g (\frac{a^{2}}{b}) + l o g (\frac{a^{3}}{b^{2}}) + l o g (\frac{a^{4}}{b^{3}}) + \dots \infty

a_{n} = \frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - . . . + {(- 1)}^{n - 1} \cdot {(\frac{3}{4})}^{n}

b_{n} = 1 - a_{n}

b_{n} > a_{n} \forall n > n_{0},

a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + . . . + (- 1)^{n - 1} (\frac{3}{4})^{n}

b_{n} = 1 - a_{n}

n_{0}

b_{n} > a_{n} \forall n > n_{0}

n \geq 4

a_{n} = n \sum j = 1 n \sum k = j (\begin{matrix} n k \end{matrix}) (\begin{matrix} k j \end{matrix})

b_{n} = a_{n} + 2^{n + 1}

b_{n + 1} = 5 b_{n} - 6 b_{n - 1}

n \geq 2

a_{n} = 5 a_{n} - 6 a_{n}

n \geq 2

Join BYJU'S Learning Program