Product Rule of the Derivatives

Formula

$\dfrac{d}{dx}{\, \Big({f{(x)}}.{g{(x)}}\Big)}$ $\,=\,$ ${f{(x)}}{\dfrac{d}{dx}{g{(x)}}}$ $+$ ${g{(x)}}{\dfrac{d}{dx}{f{(x)}}}$

Introduction

$f{(x)}$ and $g{(x)}$ are two differential functions in terms of $x$. The differentiation of product of them with respect to $x$ is written in the following mathematical form in calculus.

$\dfrac{d}{dx}{\, \Big({f{(x)}}.{g{(x)}}\Big)}$

The derivative of product can be calculated by the sum of product of first function and derivative of second function and product of second function and derivative of first function.

$\dfrac{d}{dx}{\, \Big({f{(x)}}.{g{(x)}}\Big)}$ $\,=\,$ ${f{(x)}}{\dfrac{d}{dx}{g{(x)}}}$ $+$ ${g{(x)}}{\dfrac{d}{dx}{f{(x)}}}$

This equality property is used in differential calculus for finding the differentiation of product of two or more functions.

The derivative of product rule is also simply written as $uv$ rule in calculus. It is expressed in terms of $u$ and $v$ by taking $f{(x)} = u$ and $g{(x)} = v$. It is called Leibniz’s notation for product rule.

Leibniz’s notation

$(1) \,\,\,$ $\dfrac{d}{dx}{\, (u.v)}$ $\,=\,$ $u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}$

Differentials notation

$(2) \,\,\,$ ${d}{\, (u.v)}$ $\,=\,$ $u.{dv}$ $+$ $v.{du}$

Proof

There are two differential mathematical approaches to derive the product rule of differentiation in calculus.

$(1) \,\,\,$ Learn proof for derivative product rule by the definition of the derivative in limiting operation form.

$(2) \,\,\,$ Learn proof for differentiation product rule by logarithmic system and chain rule.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved