The nth derivative of 1/(ax+b) is equal to (-1)^{n}n!a^{n}/(ax+b)^{n+1}. The n^{th} derivative of 1/(ax+b) is denoted by $\dfrac{d^n}{dx^n}\left( \dfrac{1}{ax+b}\right)$ and its formula is given below:

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

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

**Question:** What is the nth Derivative of $\dfrac{1}{ax+b}$?

**Answer:**

Let us put

y = $\dfrac{1}{x+b}$ = (ax+b)^{-1}.

Using the power rule $\dfrac{d}{dx}\left( x^n\right)=nx^{n-1}$ together of the chain rule of differentiation, the first derivative of $\frac{1}{ax+b}$ is given by

y_{1} = -1 ⋅ (ax+b)^{-1-1} $\frac{d}{dx}$(ax+b)

⇒ y_{1} = -1 ⋅ (ax+b)^{-2} ⋅ a

In a similar way, the second and the third derivatives of 1/(ax+b) are respectively given as follows:

y_{2} = (-1⋅-2) ⋅ (ax+b)^{-3} ⋅ a^{2}y _{3} = (-1⋅-2⋅-3) ⋅ (ax+b)^{-4} ⋅ a^{3} |

By looking at the patterns, we see that the nth^{ }derivative of $\frac{1}{ax+b}$ is equal to (-1⋅-2⋅-3 … -n) (ax+b)^{-(n+1)} ⋅ a^{n}.

So the nth derivative of $\frac{1}{ax+b}$ is equal to $\dfrac{(-1)^n n! a^n}{(ax+b)^{n+1}}$.

**Also Read:** nth derivative of x^{n}

What is the nth derivative of 1/x?

## Question-Answer

**Question 1:** If y = $\frac{1}{2x+1}$, then find y_{4}.

**Answer:**

From above, the nth derivative of y=1/(ax+b), denoted by y_{n}, is given by

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

Thus, to obtain the fourth order derivative y_{4} we need to put a=2, b=1 and n=4 above. Therefore, we obtain that

$y_4 = \dfrac{(-1)^4 4! 2^4}{(2x+1)^{4+1}}$ $= \dfrac{24 \times 16}{(2x+1)^5}$ $= \dfrac{384}{(2x+1)^5}$.

Hence, if y=1/(2x+1), then y_{4} = 384/(2x+1)^{5}. That is, the fourth order derivative of 1/(2x+1) is equal to 384/(2x+1)^{5}.

## FAQs

### Q1: What is nth Derivative of 1/(ax+b)?

**Answer:** The nth derivative of 1/(ax+b) is equal to (-1)^{n}n!a^{n}/(ax+b)^{n+1}.