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

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

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

The derivative of quotient can be calculated by the quotient of subtraction of product of first function and derivative of second function from product of second function and derivative of first function by the square of second function.

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

In differential calculus, the equality property is used for finding the differentiation of quotient of two functions.

The derivative of quotient rule is simply written as $u/v$ rule in calculus by taking $u = f{(x)}$ and $v = g{(x)}$.

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

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

Learn proof for derivative quotient rule by the definition of the derivative in limiting operation form.

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