>Derivation>How to Differentiate Functions Using the Quotient Rule

How to Differentiate Functions Using the Quotient Rule

The quotient rule is the rule that tells you how to take the derivative of a function that is a ratio of two differentiable functions.

Formula

The Quotient Rule

{(\frac{u}{v})}^{'} = \frac{u^{'} v - u v^{'}}{v^{2}}

where $u = u (x)$ and $v = v (x)$

Note! Sometimes you get lucky, and it’s possible to simplify the answer. Most of the time that’s not possible though, and you can just leave the fraction as it is.

Example 1

Differentiate the expression $\frac{2 x + 1}{e^{x}}$

Here, $u = 2 x + 1$ and $v = e^{x}$ . You get $u^{'} = 2$ and $v^{'} = e^{x}$ , which gives you

\begin{array}{l} {(\frac{2 x + 1}{e^{x}})}^{'} & = \frac{{(2 x + 1)}^{'} \cdot e^{x} - (2 x + 1) \cdot {(e^{x})}^{'}}{{(e^{x})}^{2}} \\ = \frac{2 \cdot e^{x} - (2 x + 1) \cdot e^{x}}{{(e^{x})}^{2}} \\ = \frac{2 e^{x} - 2 x e^{x} - e^{x}}{{(e^{x})}^{2}} \\ = \frac{e^{x} - 2 x e^{x}}{{(e^{x})}^{2}} \\ = \frac{e^{x} (1 - 2 x)}{{(e^{x})}^{2}} \\ = \frac{1 - 2 x}{e^{x}} . \end{array}

\begin{array}{l} {(\frac{2 x + 1}{e^{x}})}^{'} & = \frac{{(2 x + 1)}^{'} \cdot e^{x} - (2 x + 1) \cdot {(e^{x})}^{'}}{{(e^{x})}^{2}} \\ = \frac{2 \cdot e^{x} - (2 x + 1) \cdot e^{x}}{{(e^{x})}^{2}} \\ = \frac{2 e^{x} - 2 x e^{x} - e^{x}}{{(e^{x})}^{2}} \\ = \frac{e^{x} - 2 x e^{x}}{{(e^{x})}^{2}} \\ = \frac{e^{x} (1 - 2 x)}{{(e^{x})}^{2}} \\ = \frac{1 - 2 x}{e^{x}} \end{array}

Example 2

Differentiate the expression $\frac{3 x^{3} - 2 x^{2} + 7}{x - 1}$

Here, you have $u = 3 x^{3} - 2 x^{2} + 7$ and $v = x - 1$ . That means $u^{'} = 9 x^{2} - 2$ and $v^{'} = 1$ , and the derivative is

\begin{array}{l} {(\frac{3 x^{3} - 2 x^{2} + 7}{x - 1})}^{'} \\ = \frac{1}{{(x - 1)}^{2}} ({(3 x^{3} - 2 x^{2} + 7)}^{'} \cdot (x - 1) \\ - (3 x^{3} - 2 x^{2} + 7) {(x - 1)}^{'}) \\ = \frac{(9 x^{2} - 4 x) \cdot (x - 1) - (3 x^{3} - 2 x^{2} + 7) \cdot 1}{{(x - 1)}^{2}} \\ = \frac{9 x^{3} - 9 x^{2} - 4 x^{2} + 4 x - 3 x^{3} + 2 x^{2} - 7}{{(x - 1)}^{2}} \\ = \frac{6 x^{3} - 11 x^{2} + 4 x - 7}{{(x - 1)}^{2}} \end{array}

\begin{array}{l} {(\frac{3 x^{3} - 2 x^{2} + 7}{x - 1})}^{'} & = \frac{{(3 x^{3} - 2 x^{2} + 7)}^{'} \cdot (x - 1) - (3 x^{3} - 2 x^{2} + 7) {(x - 1)}^{'}}{{(x - 1)}^{2}} \\ = \frac{(9 x^{2} - 4 x) \cdot (x - 1) - (3 x^{3} - 2 x^{2} + 7) \cdot 1}{{(x - 1)}^{2}} \\ = \frac{9 x^{3} - 9 x^{2} - 4 x^{2} + 4 x - 3 x^{3} + 2 x^{2} - 7}{{(x - 1)}^{2}} \\ = \frac{6 x^{3} - 11 x^{2} + 4 x - 7}{{(x - 1)}^{2}} . \end{array}

x = 1

is not a root of the numerator, you can’t simplify the expression.

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