The nth derivative of 1/x is denoted by $\frac{d^n}{dx^n}(\frac{1}{x})$ and it is equal to (-1)^{n}n!/x^{n+1}. The nth derivative of 1/(ax+b) is denoted by $\frac{d^n}{dx^n}(\frac{1}{ax+b})$ and it is equal to (-1)^{n}a^{n}n!/(ax+b)^{n+1}.

So the n-th derivative formulas of 1/x and logx are given as follows:

- $\dfrac{d^n}{dx^n} \left(\dfrac{1}{x} \right)$ = $\dfrac{(-1)^n n!}{x^{n+1}}$.
- $\dfrac{d^n}{dx^n} \left(\dfrac{1}{ax+b} \right)$ = $\dfrac{(-1)^n a^nn!}{(ax+b)^{n+1}}$.

Table of Contents

## nth Derivative of 1/x

**Question:** What is the nth derivative of 1/x?

**Solution:**

Let y = $\dfrac{1}{x}$.

So y = x^{-1}.

Differentiating with respect to x, we get that

y_{1} = -1 ⋅ x^{-1-1}

⇒ y_{1} = -1 ⋅ x^{-2}

Again differentiating y_{1} with respect to x, we obtain that

y_{2} = -1 ⋅ -2 ⋅ x^{-2-1} ⇒ y_{2} = (-1 ⋅ -2) x^{-3}

In the same way, y _{3} = (-1 ⋅ -2 ⋅ -3) x^{-4}y _{4} = (-1 ⋅ -2⋅ -3⋅ -4) x^{-5} |

Continuing in the same fashion, we deduce that the nth derivative of 1/x is equal to y_{n} = (-1 ⋅ -2⋅ -3⋅⋅⋅ -n) x^{-(n+1)} = (-1)^{n}n!/x^{n+1}.

**Also Read:** nth Derivative of sinx and cosx

## nth Derivative of 1/(ax+b)

**Question:** What is the nth derivative of 1/(ax+b)?

**Solution:**

Let y = $\dfrac{1}{a+bx}$.

So y = (ax+b)^{-1}.

Let us differentiate y with respect to x. Thus, we get

y_{1} = -1 ⋅ a(ax+b)^{-1-1}

⇒ y_{1} = -1 ⋅ a(ax+b)^{-2}

Now differentiate y_{1} with respect to x. So

y_{2} = -1 ⋅ -2 ⋅ a⋅a(ax+b)^{-2-1} ⇒ y_{2} = (-1 ⋅ -2) a^{2}(ax+b)^{-3}

In the same way, we obtain the following derivatives y _{3} = (-1 ⋅ -2 ⋅ -3) a^{3}(ax+b)^{-4}y _{4} = (-1 ⋅ -2⋅ -3⋅ -4) a^{4}(ax+b)^{-5} |

If we proceed in the same fashion, we deduce that the nth derivative of 1/(ax+b) is equal to y_{n} = (-1 ⋅ -2⋅ -3⋅⋅⋅ -n) a^{n}(ax+b)^{-(n+1)} = (-1)^{n}a^{n}n!/(ax+b)^{n+1}.

So the n times differentiation of 1/(ax+b) with respective to x is given by

$\dfrac{(-1)^n a^n n!}{(ax+b)^{n+1}}$.

## FAQs

**Q1: If y=1/x then find the n-th derivative y**

_{n}.Answer: If y=1/x, then its nth derivative y_{n} = (-1)^{n}n!/x^{n+1}.

**Q2: If y=1/(ax+b) then find the nth derivative y**

_{n}.Answer: If y=1/(ax+b), then its nth derivative y_{n} = (-1)^{n}a^{n}n!/(ax+b)^{n+1}.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.