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Reciprocal rule of Derivatives

Formula

$\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\,\dfrac{1}{\Big(f(x)\Big)^2} \times \dfrac{d}{dx}{\Big(f(x)\Big)}$

Introduction

In differential calculus, the functions in reciprocal form are appeared sometimes. It is difficult to perform the differentiation of such functions with derivative rules. So, a special rule is required to find the derivative of a reciprocal function and it is called the reciprocal rule of the derivatives.

Let $f(x)$ be a function in terms of $x$ and its multiplicative inverse is written as follows in mathematics.

$\dfrac{1}{f(x)}$

The derivative of the reciprocal of the function $f(x)$ is written in the following mathematical form.

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$

The derivative of the multiplicative inverse of the function $f(x)$ with respect to $x$ is equal to negative product of the quotient of one by square of the function and the derivative of the function with respect to $x$.

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\,\dfrac{1}{\Big(f(x)\Big)^2} \times \dfrac{d}{dx}{\Big(f(x)\Big)}$

This mathematical relation is called the reciprocal rule of the differentiation.

In differential calculus, the derivative of the function with respect to $x$ is simplify written as $f'(x)$. Hence, the reciprocal rule of the derivatives can also written as follows.

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\,\dfrac{1}{\Big(f(x)\Big)^2} \times f'(x)$

Example

Evaluate $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$

Now, use the reciprocal rule of the derivatives to find its differentiation.

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$ $\,=\,$ $-\dfrac{1}{(2x+3)^2} \times \dfrac{d}{dx}{(2x+3)}$

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$ $\,=\,$ $-\dfrac{1}{(2x+3)^2} \times \Big(\dfrac{d}{dx}{(2x)}+\dfrac{d}{dx}{(3)}\Big)$

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$ $\,=\,$ $-\dfrac{1}{(2x+3)^2} \times (2+0)$

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$ $\,=\,$ $-\dfrac{1}{(2x+3)^2} \times (2)$

$\implies$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$ $\,=\,$ $-\dfrac{1 \times 2}{(2x+3)^2}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{2x+3}\bigg)}$ $\,=\,$ $-\dfrac{2}{(2x+3)^2}$

Proof

Learn how to derive the reciprocal rule of the differentiation in mathematics.

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