A derivative is defined as the rate of change of a function with respect to its given domain or input over a specified limit. In the study of Calculus, the derivative of a function can be found with the help of several tools including the Quotient Rule. The Quotient Rule is most appropriately used when finding the derivative of a function that is a ratio or quotient of two polynomials.

The Quotient Rule states:

Let F(x) = j(x)/k(x); where j(x) and k(x) are functions for which there is a derivative and by definition polynomials.

F’(x) = d/dx[j(x)/k(x)] = [ k(x)[j’(x)] – j(x)[k’(x)] ] / (k(x))^{2}

Stating this in words, the Quotient Rule derivation says that the derivative of function F (a quotient with polynomial j as its numerator and polynomial k as its denominator each with respect to x) is the difference of the denominator multiplied by the derivative of the numerator minus the numerator multiplied by the derivative of the denominator all of which is divided by the denominator squared.

Using a little bit of algebra and the use of the Power Rule to differentiate the individual polynomials make the Quotient Rule a simple yet powerful tool for finding derivatives.

An example will help to illustrate how this works.

Let’s say we want to find the derivative of f(x) = (x^{2} + 6) / (x^{3} – 5x^{2}).

So we will rewrite this function in the form outlined as stated above.

f’(x) = [ (x^{3} – 5x^{2}) d/dx[x^{2} + 6] – (x^{2} + 6) d/dx[x^{3} – 5x^{2}] ] / (x^{3} – 5x^{2})^{2}

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Now we will use the Power Rule to find the derivatives of polynomials j(x) and k(x) and then substitute those back in to the formula.

f’(x) = [ (x^{3} – 5x^{2}) (2x) – (x^{2} + 6) (3x^{2} – 10x) ] / (x^{3} – 5x^{2})^{2}

Using some simple yet involved algebraic operations will allow us to complete this derivative. For simplicity, we will look at the numerator and denominator individually reducing each down to its simplest term.

Numerator: [ (2x^{4} – 10x^{3}) – (3x^{4}–10x^{3} + 18x^{2} – 60x) ] = –x^{4} – 18x^{2} + 60x

Denominator: (x^{3} – 5x^{2}) (x^{3} – 5x^{2}) = x^{6} – 5x^{5} – 5x^{5} + 25x^{4 }= x^{6 }– 10x^{5 }+ 25x^{4}

So f’(x) = (–x^{4} – 18x^{2} + 60x) / (x^{6 }–10x^{5 }+ 25x^{4})

The Quotient Rule can also be used in conjunction with other rules of differentiation such as the Power, Product and Chain rules for more complex derivatives as the situation allows.

Resource:

Stewart, James. (2001). Calculus Concepts And Contexts (5^{th} Ed.). Pacific Grove, CA: Wadsworth Group.

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