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

$f{(x)}$ 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 Leibnizâ€™s notation for product rule.

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

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

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.

Latest Math Topics

Apr 18, 2022

Apr 14, 2022

Apr 05, 2022

Mar 18, 2022

Mar 05, 2022

Latest Math Problems

Apr 06, 2022

Mar 22, 2022

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved