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?

Solution:

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 y_n = (-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)?

Solution:

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 y_n = (-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

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

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