Introduction
In calculus, derivatives are an extremely useful tool that are used in a variety of problems. In this article, we will specifically learn about taking the derivatives of fractions.
Derivatives of Fractions
A fraction typically appears as the ratio of two functions in the form $\dfrac{f(x)}{g(x)}$, where \( f(x) \) is the numerator and \( g(x) \) is the denominator.
To find the derivative of such an expression, we use the Quotient Rule, which says
\[\begin{align*} \dfrac{d}{dx} \left(\dfrac{f(x)}{g(x)}\right) = \dfrac{g(x)f'(x) – f(x)g'(x)}{g(x)^2}. \end{align*}\]
The Quotient Rule is a notation-heavy formula that is often difficult to memorize. A helpful mnemonic is “Low Dee High, High Dee Low, all over Low Low.” To clarify, the denominator is “Low,” the numerator is “High,” and the “Dee” represents the derivative function.
Examples
Suppose we want to take the derivative of $\dfrac{x^2}{x + 1}$. Since we have a fraction, this reminds us of using the Quotient Rule. Notice that
$$ \dfrac{x^2}{x + 1} = \dfrac{f(x)}{g(x)}, $$
so it becomes clear that $f(x) = x^2$ and $g(x) = x + 1$.
Before applying the Quotient Rule, a helpful tip is to take the derivative of $f(x)$ and $g(x)$ beforehand. Using the Power Rule, we have
\[\begin{align*} &f(x) = x^2 \quad\:\: g(x) = x + 1 \\ &f'(x) = 2x \quad g'(x) = 1. \end{align*}\]
Now, we simply plug into our formula:
\[\begin{align*} \dfrac{d}{dx} \left(\dfrac{f(x)}{g(x)}\right) &= \dfrac{g(x)f'(x) – f(x)g'(x)}{g(x)^2} \\ &= \dfrac{(x + 1)(2x) – (x^2)(1)}{(x + 1)^2} \\ &= \dfrac{2x^2 + 2x – x^2}{x^2 + 2x + 1} \\ &= \boxed{\dfrac{x^2 + 2x}{x^2 + 2x + 1}}. \end{align*}\]
Let’s try another example. Suppose we want to find the derivative of $\dfrac{\sin x}{\cos x}$. Like before, notice that
$$\dfrac{\sin x}{\cos x} = \dfrac{f(x)}{g(x)},$$
so it becomes clear that $f(x) = \sin x$ and $g(x) = \cos x$. We can organize the derivatives of $f(x)$ and $g(x)$ before we apply the Quotient Rule:
\[\begin{align*} &f(x) = \sin x \quad\:\: g(x) = \cos x \\ &f'(x) = \cos x \quad g'(x) = -\sin x. \end{align*}\]
Notice that we had to use the identity $\dfrac{d}{dx}(\sin x) = \cos x$ and $\dfrac{d}{dx}(\cos x) = -\sin x$.
Now, we can plug this into our Quotient Rule formula:
\[\begin{align*} \dfrac{d}{dx} \left(\dfrac{f(x)}{g(x)}\right) &= \dfrac{g(x)f'(x) – f(x)g'(x)}{g(x)^2} \\ &= \dfrac{(\cos x)(\cos x) – (\sin x)(-\sin x)}{(-\sin x)^2} \\ &= \dfrac{\cos^2x + \sin^2x}{\sin^2x} \\ &= \dfrac{1}{\sin^2x} \\ &= \boxed{\sec^2x}. \end{align*}\]
Again, we resorted to using the identity $\sin^2x + \cos^2x = 1$ to simplify our expression. For information on why this is true, check out other blogs about trigonometric identities on iacedcalculus.com.
Summary
When applying the Quotient Rule, it will always be helpful to follow these three steps:
- Identify $f(x)$ and $g(x)$.
- Find $f'(x)$ and $g'(x)$ in order to organize your information before applying the Quotient Rule.
- Apply the Quotient Rule and simplify the expression.
This rule is a powerful tool in calculus for handling derivatives of functions in fraction form, and mastering it will enable you to solve many problems derivatives of complex functions.
