The nth derivative of 1/x is equal to (-1)^{n}n!/x^{n+1}. This is obtained by repeatedly using the power rule of differentiation.

The n^{th} derivative of 1/x is denoted by $\dfrac{d^n}{dx^n}\left( \dfrac{1}{x}\right)$, and its formula is given as follows:

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

## nth Derivative of 1/x

**Question:** How to find nth Derivative of 1/x?

**Answer:**

To find the n^{th }derivative of 1/x, let us put y = 1/x. We need to find y_{n}, the nth derivative of y. We have:

y = x^{-1}.

By power rule $\dfrac{d}{dx}\left( x^n\right)=nx^{n-1}$, the first derivative of 1/x is given by

y_{1} = -1 ⋅ x^{-1-1} = -1 ⋅ x^{-2}.

The second derivative of y is obtained by differentiating y_{1} with respect to x. That is,

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

In a similar way, the third and the fourth order derivatives of 1/x are respectively given as follows:

y_{3} = -1⋅-2⋅-3 ⋅ x^{-3-1} = (-1⋅-2⋅-3) x^{-4}.

y_{4} = -1⋅-2⋅-3⋅-4 ⋅ x^{-4-1} = (-1⋅-2⋅-3⋅-4) x^{-5}.

By looking at the patterns, we see that the n^{th }derivative of 1/x is equal to (-1⋅-2⋅-3⋅-4 … -n) x^{-(n+1)} = (-1)^{n}n!/x^{n+1}.

**Conclusion:** Therefore, the n^{th} derivative of 1/x is equal to $\dfrac{(-1)^n n!}{x^{n+1}}$.

## Question-Answer

**Question 1:** If y=1/x, then find y_{5}.

**Answer:**

From above, we know that the nth derivative of 1/x is equal to (-1)^{n}n!/x^{n+1}. That is, y_{n} = $\dfrac{(-1)^n n!}{x^{n+1}}$. Thus, to obtain the fifth derivative y_{5}, let us put n=5.

So $y_5 = \dfrac{(-1)^5 5!}{x^{5+1}}$ $= -\dfrac{120}{x^6}$.

Hence, if y=1/x, then y_{5} = -120/x^{6}. That is, the fifth order derivative of 1/x is equal to -120/x^{6}.

Similarly, the sixth derivative of 1/x is equal to $\dfrac{(-1)^6 6!}{x^{6+1}}$ = 720/x^{7}, that is, y_{6} = 720/x^{7}.

**Also Read:** Derivative of y=sin(x+y)

## FAQs

### Q1: What is nth Derivative of 1/x?

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