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 $x$.

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

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 $x$.

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

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.

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

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.

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

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

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

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.

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

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

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

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

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