Proof of Product Rule of Differentiation
Fact-checked:
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.
$\dfrac{d}{dx}{\Big({f(x)}.{g(x)}\Big)}$
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.
$(2) \,\,\,\,\,\,$ $d{(u.v)}$ $\,=\,$ $v.du$ $+$ $u.dv$
