What is the nth Derivative of 1/x? [Solved]

The nth derivative of 1/x is equal to (-1)ⁿn!/xⁿ⁺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

y₁ = -1 ⋅ x^-1-1 = -1 ⋅ x^-2.

The second derivative of y is obtained by differentiating y₁ with respect to x. That is,

y₂ = -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:

y₃ = -1⋅-2⋅-3 ⋅ x^-3-1 = (-1⋅-2⋅-3) x^-4.

y₄ = -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)ⁿn!/xⁿ⁺¹.

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

Derivative of x root(x)

Question-Answer

Question 1: If y=1/x, then find y₅.

Answer:

From above, we know that the nth derivative of 1/x is equal to (-1)ⁿn!/xⁿ⁺¹. That is, y_n = $\dfrac{(-1)^n n!}{x^{n+1}}$. Thus, to obtain the fifth derivative y₅, 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 y₅ = -120/x⁶. That is, the fifth order derivative of 1/x is equal to -120/x⁶.

Similarly, the sixth derivative of 1/x is equal to $\dfrac{(-1)^6 6!}{x^{6+1}}$ = 720/x⁷, that is, y₆ = 720/x⁷.

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

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