Math Doubts

Proof of Quotient rule of Differentiation

$f{(x)}$ and $g{(x)}$ are two differential functions in terms of $x$. The quotient of them is written as $\dfrac{f{(x)}}{g{(x)}}$ in mathematics and the derivative of quotient of them with respect to $x$ is written in the following mathematical form.

$\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$

Take $q{(x)} = \dfrac{f{(x)}}{g{(x)}}$

$\implies$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\dfrac{d}{dx}{\, q{(x)}}$

Use Definition of the Derivative

According to the definition of the derivative, the derivative of the quotient of two differential functions can be written in the form of limiting operation for finding the differentiation of quotient by first principle.

$\dfrac{d}{dx}{\, q{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{q{(x+\Delta x)}-q{(x)}}{\Delta x}}$

Take $\Delta x = h$ and replace the $\Delta x$ by $h$ in the right-hand side of the equation.

$\implies$ $\dfrac{d}{dx}{\, q{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{q{(x+h)}-q{(x)}}{h}}$

We have taken that $q{(x)} = \dfrac{f{(x)}}{g{(x)}}$, then $q{(x+h)} = \dfrac{f{(x+h)}}{g{(x+h)}}$. Now, replace the functions $q{(x+h)}$ and $q{(x)}$ by their actual values.

$\implies$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{f{(x+h)}}{g{(x+h)}}-\dfrac{f{(x)}}{g{(x)}}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\dfrac{{f{(x+h)}}{g{(x)}}-{g{(x+h)}}{f{(x)}}}{{g{(x+h)}}{g{(x)}}}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{f{(x+h)}}{g{(x)}}-{g{(x+h)}}{f{(x)}}}{h \times {g{(x+h)}}{g{(x)}}}}$

A small acceptable Adjustment

Now, add and subtract $f{(x)}g{(x)}$ in the numerator of the function for factoring the mathematical expression.

$\implies$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{f{(x+h)}}{g{(x)}}-{g{(x+h)}}{f{(x)}}+{f{(x)}}{g{(x)}}-{f{(x)}}{g{(x)}}}{h \times {g{(x+h)}}{g{(x)}}}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{f{(x+h)}}{g{(x)}}-{f{(x)}}{g{(x)}}-{g{(x+h)}}{f{(x)}}+{f{(x)}}{g{(x)}}}{h \times {g{(x+h)}}{g{(x)}}}}$

Simplify the mathematical expression

In the numerator, $g{(x)}$ is a common factor in the first two terms and $f{(x)}$ is a common factor in the remaining two terms. So, take them common to take a first step in simplifying this mathematical expression.

$\implies$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{g{(x)}}\Big(f{(x+h)}-f{(x)}\Big)-{f{(x)}}\Big(g{(x+h)}-g{(x)}\Big)}{h \times {g{(x+h)}}{g{(x)}}}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{{g{(x)}}\Big(f{(x+h)}-f{(x)}\Big)-{f{(x)}}\Big(g{(x+h)}-g{(x)}\Big)}{h}}$ $\times$ $\dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg]$

Use product rule of limits for evaluating limit of product of two functions by evaluating product of their limits.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{g{(x)}}\Big(f{(x+h)}-f{(x)}\Big)-{f{(x)}}\Big(g{(x+h)}-g{(x)}\Big)}{h}} \normalsize \Bigg)$ $\times$ $\Bigg( \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg)}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{{g{(x)}}\Big(f{(x+h)}-f{(x)}\Big)}{h}-\dfrac{{f{(x)}}\Big(g{(x+h)}-g{(x)}\Big)}{h}\Bigg]} \normalsize \Bigg)$ $\times$ $\Bigg( \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg)}$

Now, use difference rule of limits for calculating limit of difference of functions by difference of their limits.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{g{(x)}}\Big(f{(x+h)}-f{(x)}\Big)}{h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{{f{(x)}}\Big(g{(x+h)}-g{(x)}\Big)}{h}} \normalsize \Bigg)$ $\times$ $\Bigg( \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg)}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[g{(x)} \times \dfrac{f{(x+h)}-f{(x)}}{h}\Bigg]}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[ f{(x)} \times \dfrac{g{(x+h)}-g{(x)}}{h}\Bigg]} \normalsize \Bigg)$ $\times$ $\Bigg( \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg)}$

Try product rule of limits and find limit of product of functions in each term of the first factor of the expression.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize g{(x)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize f{(x)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}} \normalsize \Bigg)$ $\times$ $\Bigg( \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg)}$

Evaluate Limits of the functions

$\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\Bigg(\displaystyle \large \lim_{h \,\to\, 0}{\normalsize g{(x)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize f{(x)}}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}} \normalsize \Bigg)$ $\times$ $\Bigg( \displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{1}{{g{(x+h)}}{g{(x)}}} \Bigg)}$

Evaluate the limit of first factor of each term in the first factor and second factor by the direct substitution method.

$=\,\,\,$ $\Bigg(g{(x)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $-$ $f{(x)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}} \normalsize \Bigg)$ $\times$ $\Bigg( \dfrac{1}{{g{(x+0)}}{g{(x)}}}\Bigg)$

$=\,\,\,$ $\Bigg(g{(x)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $-$ $f{(x)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}} \normalsize \Bigg)$ $\times$ $\Bigg( \dfrac{1}{{g{(x)}}{g{(x)}}}\Bigg)$

$=\,\,\,$ $\Bigg(g{(x)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $-$ $f{(x)}$ $\times$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}} \normalsize \Bigg)$ $\times$ $\Bigg(\dfrac{1}{g{(x)}^2}\Bigg)$

The limit of the function as $h$ approaches $0$ is derivative of the respective function as per the definition of the derivative in limiting operation.

$\implies$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\Bigg(g{(x)}$ $\times$ $\dfrac{d}{dx}{\, f{(x)}}$ $-$ $f{(x)}$ $\times$ $\dfrac{d}{dx}{\, g{(x)}} \Bigg)$ $\times$ $\Bigg(\dfrac{1}{g{(x)}^2}\Bigg)$

$\implies$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\dfrac{g{(x)} \times \dfrac{d}{dx}{\, f{(x)}} -f{(x)} \times \dfrac{d}{dx}{\, g{(x)}}}{g{(x)}^2}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{f{(x)}}{g{(x)}}\Bigg)}$ $\,=\,$ $\dfrac{g{(x)}\dfrac{d}{dx}{\, f{(x)}} -f{(x)}\dfrac{d}{dx}{\, g{(x)}}}{g{(x)}^2}$

This property is called the quotient rule of derivatives and it is used to find the differentiation of quotient of any two differential functions.

Alternative forms

The quotient rule of differentiation is written in two different forms by taking $u = f{(x)}$ and $v = g{(x)}$.

Leibniz’s notation

$(1) \,\,\,$ $\dfrac{d}{dx}{\, \Bigg(\dfrac{u}{v}\Bigg)}$ $\,=\,$ $\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}$

Differentials notation

$(2) \,\,\,$ ${d}{\, \Bigg(\dfrac{u}{v}\Bigg)}$ $\,=\,$ $\dfrac{v{du}-u{dv}}{v^2}$



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