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Proof of Product Rule of Differentiation

The proof of derivative product rule can be derived in calculus by first principle as per definition of the derivative. It can also be derived in another mathematical approach.

$f{(x)}$ and $g{(x)}$ are two functions in terms of $x$ and their product is equal to $f{(x)}.g{(x)}$. The differentiation of the product with respect to $x$ is written in mathematics in the following way.


Step for deriving the product rule

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

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

Differentiate this mathematical 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 product of the functions can be separated by the logarithms.

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

Use the product rule of logarithms and and separate both functions as sum of their logs.

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

Differentiate the Logarithmic Equation

The logarithmic equation is in terms of $x$. So, differentiate both sides of 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.

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

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

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

Simplify the Differential Equation

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