Find nth Derivative of 1/(ax+b)

The nth derivative of 1/(ax+b) is equal to (-1)ⁿn!aⁿ/(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

y₁ = -1 ⋅ (ax+b)^-1-1 $\frac{d}{dx}$(ax+b)

⇒ y₁ = -1 ⋅ (ax+b)^-2 ⋅ a

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

y2 = (-1⋅-2) ⋅ (ax+b)-3 ⋅ a2 y3 = (-1⋅-2⋅-3) ⋅ (ax+b)-4 ⋅ a3

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) ⋅ aⁿ.

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 xⁿ

What is the nth derivative of 1/x?

Question-Answer

Question 1: If y = $\frac{1}{2x+1}$, then find y₄.

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 y₄ 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 y₄ = 384/(2x+1)⁵. That is, the fourth order derivative of 1/(2x+1) is equal to 384/(2x+1)⁵.

FAQs