nth derivative of 1/x and 1/(ax+b)

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

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

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

nth Derivative of 1/x

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

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

So y = x-1.

Differentiating with respect to x, we get that

y1 = -1 ⋅ x-1-1

⇒ y1 = -1 ⋅ x-2

Again differentiating y1 with respect to x, we obtain that

y2 = -1 ⋅ -2 ⋅ x-2-1 ⇒ y2 = (-1 ⋅ -2) x-3

In the same way,

y3 = (-1 ⋅ -2 ⋅ -3) x-4

y4 = (-1 ⋅ -2⋅ -3⋅ -4) x-5

Continuing in the same fashion, we deduce that the nth derivative of 1/x is equal to yn = (-1 ⋅ -2⋅ -3⋅⋅⋅ -n) x-(n+1) = (-1)nn!/xn+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)?

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

So y = (ax+b)-1.

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

y1 = -1 ⋅ a(ax+b)-1-1

⇒ y1 = -1 ⋅ a(ax+b)-2

Now differentiate y1 with respect to x. So

y2 = -1 ⋅ -2 ⋅ a⋅a(ax+b)-2-1 ⇒ y2 = (-1 ⋅ -2) a2(ax+b)-3

In the same way, we obtain the following derivatives

y3 = (-1 ⋅ -2 ⋅ -3) a3(ax+b)-4

y4 = (-1 ⋅ -2⋅ -3⋅ -4) a4(ax+b)-5

If we proceed in the same fashion, we deduce that the nth derivative of 1/(ax+b) is equal to yn = (-1 ⋅ -2⋅ -3⋅⋅⋅ -n) an(ax+b)-(n+1) = (-1)nann!/(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 yn.

Answer: If y=1/x, then its nth derivative yn = (-1)nn!/xn+1.

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

Answer: If y=1/(ax+b), then its nth derivative yn = (-1)nann!/(ax+b)n+1.

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