Math Doubts

Proof of Product rule of Differentiation


General form
$\dfrac{d}{dx}{\Big({f(x)}.{g(x)}\Big)}$ $\,=\,$ ${g(x)}\dfrac{d{f(x)}}{dx}$ $+$ ${f(x)}\dfrac{d{g(x)}}{dx}$
Leibniz’s notation
$\dfrac{d}{dx}{(u.v)}$ $\,=\,$ $v.\dfrac{du}{dx}$ $+$ $u.\dfrac{dv}{dx}$
Differentials Notation
$d{(u.v)}$ $\,=\,$ $v.du$ $+$ $u.dv$

$f{(x)}$ and $g{(x)}$ are two functions and the product of them is equal to $f{(x)}.g{(x)}$. The two functions $f{(x)}$ and $g{(x)}$ are actually functions in terms of $x$. So, it should be differentiated with respect to $x$.


Essential steps for deriving product rule

Assume, the product of the two functions $f{(x)}$ and $g{(x)}$ is equal to $y$.

$y$ $\,=\,$ ${f(x)}.{g(x)}$

Differentiate the product of the equation with respect to $x$.

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

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

The derivative of $y$ with respect to $x$ is equal to the derivative of product of the functions $f{(x)}$ and $g{(x)}$ with respect to $x$.

Separate the product of the functions

$y$ $\,=\,$ ${f(x)}.{g(x)}$

The function is form of the product of two functions so that it is not possible to differentiate the function easily. Therefore, try to divide product of the functions separately. It is only possible by the logarithm system.

Now, take natural logarithm both sides of the equation.

$\implies \log_{e}{y}$ $\,=\,$ $\log_{e}{[{f(x)}.{g(x)}]}$

Use the product rule of logarithms and it separates the logarithm of product of functions as sum of their logs.

$\implies \log_{e}{y}$ $\,=\,$ $\log_{e}{[f(x)]}$ $+$ $\log_{e}{[g(x)]}$

Differentiate Logarithmic Equation

Now, differentiate the logarithmic equation with respect to $x$.

$\implies \dfrac{d}{dx}\log_{e}{y}$ $\,=\,$ $\dfrac{d}{dx}\Big[\log_{e}{f(x)}$ $+$ $\log_{e}{g(x)}\Big]$

$\implies \dfrac{d}{dx} \log_{e}{y}$ $\,=\,$ $\dfrac{d}{dx} \log_{e}{f(x)}$ $+$ $\dfrac{d}{dx} \log_{e}{g(x)}$

Each term in this logarithmic equation is formed by the composition of two functions. Hence, use chain rule to differentiate each term in the equation.

$\implies \dfrac{1}{y} \dfrac{d}{dx} y$ $\,=\,$ $\dfrac{1}{f{(x)}} \dfrac{d}{dx}{f(x)}$ $+$ $\dfrac{1}{g{(x)}} \dfrac{d}{dx}{g(x)}$

$\implies \dfrac{1}{y}\dfrac{dy}{dx}$ $\,=\,$ $\dfrac{1}{f{(x)}}\dfrac{d{f(x)}}{dx}$ $+$ $\dfrac{1}{g{(x)}}\dfrac{d{g(x)}}{dx}$

$\implies \dfrac{dy}{dx}$ $\,=\,$ $y\Bigg[\dfrac{1}{f{(x)}}\dfrac{d{f(x)}}{dx}$ $+$ $\dfrac{1}{g{(x)}}\dfrac{d{g(x)}}{dx}\Bigg]$

Simplifying the Differential Equation

It is taken that $y$ $\,=\,$ ${f(x)}.{g(x)}$. So, replace the value of $y$ by its equal value.

$\implies \dfrac{d}{dx}{\Big({f(x)}.{g(x)}\Big)}$ $\,=\,$ ${f(x).g(x)} \times \Bigg[\dfrac{1}{f{(x)}}\dfrac{d{f(x)}}{dx}$ $+$ $\dfrac{1}{g{(x)}}\dfrac{d{g(x)}}{dx}\Bigg]$

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

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

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

Therefore, it is proved that the derivative of product of two functions is equal to sum of the products one function and differentiation of second function.

Alternate forms

Gottfried Wilhelm von Leibniz, a German mathematician who simplify expressed this formula in simple notation by taking $f{(x)} \,=\, u$ and $g{(x)} \,=\, v$. Now, write the product rule of differentiation in Leibniz’s notation

Leibniz’s notation

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

Differentials notation

The product rule of differentiation can also be expressed in differential form.

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

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