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Encyclopedia>Functions>Derivation and Its Applications>Derivation>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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