2.4 First Derivatives of the Quotient of Two Polynomials

2.4 Find the Derivatives of a Quotient using Quotient Rule

Method 1

The Quotient Rule

Example:

Method 2 (Differentiate Directly)

Example:

Given that y = \frac{x^{2}}{2 x + 1}, find \frac{d y}{d x}

Solution:

\begin{array}{l} y = \frac{x^{2}}{2 x + 1} \\ \frac{d y}{d x} = \frac{(2 x + 1) (2 x) - x^{2} (2)}{{(2 x + 1)}^{2}} \\ = \frac{4 x^{2} + 2 x - 2 x^{2}}{{(2 x + 1)}^{2}} = \frac{2 x^{2} + 2 x}{{(2 x + 1)}^{2}} \end{array}

Practice 1:

Given that y = \frac{4 x^{3}}{{(5 x + 1)}^{3}}, find \frac{d y}{d x}

Solution:

\begin{array}{l} y = \frac{4 x^{3}}{{(5 x + 1)}^{3}} \\ \frac{d y}{d x} = \frac{{(5 x + 1)}^{3} (12 x^{2}) - 4 x^{3} .3 {(5 x + 1)}^{2} .5}{{[{(5 x + 1)}^{3}]}^{2}} \\ = \frac{{(5 x + 1)}^{3} (12 x^{2}) - 60 x^{3} {(5 x + 1)}^{2}}{{(5 x + 1)}^{6}} \\ = \frac{(12 x^{2}) {(5 x + 1)}^{2} [(5 x + 1) - 5 x]}{{(5 x + 1)}^{6}} \\ = \frac{(12 x^{2}) {(5 x + 1)}^{2} (1)}{{(5 x + 1)}^{6}} \\ = \frac{12 x^{2}}{{(5 x + 1)}^{4}} \end{array}