The Quotient Rule and Trig Derivatives

The quotient rule differentiates one function divided by another, like $\dfrac{x}{x^2 + 1}$ or $\dfrac{e^x}{x^2}$. It also unlocks the derivatives of the other four trig functions, since $\tan x$, $\cot x$, $\sec x$, and $\csc x$ can all be written as quotients of $\sin x$ and $\cos x$. As always in calculus, angles are in radians.

Key ideas

The quotient rule

If $f$ and $g$ are differentiable and $g(x) \ne 0$, then

$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\big[g(x)\big]^2}$$

A popular memory aid, with “hi” for the top and “lo” for the bottom: “lo d-hi minus hi d-lo, over lo squared.”

Unlike the product rule, order matters here, because of the minus sign. Always start with the derivative of the top.

When not to use it

If the denominator is a constant or a single power of $x$, it’s usually faster to rewrite and use the power rule:

• $\dfrac{x^3 + 1}{4} = \dfrac{1}{4}x^3 + \dfrac{1}{4}$, so the derivative is $\dfrac{3}{4}x^2$.
• $\dfrac{x^2 - 6}{x} = x - 6x^{-1}$, so the derivative is $1 + \dfrac{6}{x^2}$.

Derivatives of tan x and sec x

Write $\tan x = \dfrac{\sin x}{\cos x}$ and use the quotient rule:

$$\begin{aligned} \frac{d}{dx}[\tan x] &= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} \\ &= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \\ &= \frac{1}{\cos^2 x} = \sec^2 x \end{aligned}$$

The key step is the Pythagorean identity $\sin^2 x + \cos^2 x = 1$.

For $\sec x = \dfrac{1}{\cos x}$, the top is the constant $1$, whose derivative is $0$:

$$\frac{d}{dx}[\sec x] = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x \tan x$$

All six trig derivatives

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$

A pattern to help you remember: the three “co” functions (cosine, cotangent, cosecant) have derivatives with a minus sign. Practice 4 and 5 ask you to derive $\cot x$ and $\csc x$ yourself.

Worked examples

Example 1: A rational function

Differentiate $y = \dfrac{2x + 1}{x - 3}$.

Solution. Top: $f = 2x + 1$, $f' = 2$. Bottom: $g = x - 3$, $g' = 1$.

$$\begin{aligned} \frac{dy}{dx} &= \frac{(2)(x - 3) - (2x + 1)(1)}{(x - 3)^2} \\ &= \frac{2x - 6 - 2x - 1}{(x - 3)^2} \\ &= \frac{-7}{(x - 3)^2} \end{aligned}$$

Notice the brackets around $(2x + 1)$: the minus sign applies to the whole thing.

Example 2: Finding horizontal tangents

Find the points where $f(x) = \dfrac{x}{x^2 + 1}$ has a horizontal tangent.

Solution.

$$f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2}$$

A fraction is $0$ when its numerator is $0$ (and its denominator isn’t). Here the denominator is never $0$, so solve $1 - x^2 = 0$: $x = \pm 1$.

$f(1) = \dfrac{1}{2}$ and $f(-1) = -\dfrac{1}{2}$. The points are $\left(1, \dfrac{1}{2}\right)$ and $\left(-1, -\dfrac{1}{2}\right)$.

$f(x) = \dfrac{x}{x^2 + 1}$ has horizontal tangents at $\left(1, \tfrac{1}{2}\right)$ and $\left(-1, -\tfrac{1}{2}\right)$, exactly where $f'(x) = 0$.

Example 3: A tangent line with tan x

Find the equation of the tangent line to $y = \tan x$ at $x = \dfrac{\pi}{4}$.

Solution. Point: $\tan \dfrac{\pi}{4} = 1$. Slope: $\sec^2 \dfrac{\pi}{4} = \dfrac{1}{\cos^2 (\pi/4)} = \dfrac{1}{1/2} = 2$.

$$y - 1 = 2\left(x - \frac{\pi}{4}\right)$$

Example 4: From a table

Let $q(x) = \dfrac{f(x)}{g(x)}$, where $f(2) = 3$, $f'(2) = -1$, $g(2) = 4$, and $g'(2) = 2$. Find $q'(2)$.

Solution.

$$q'(2) = \frac{f'(2)g(2) - f(2)g'(2)}{\big[g(2)\big]^2} = \frac{(-1)(4) - (3)(2)}{4^2} = \frac{-10}{16} = -\frac{5}{8}$$

Common mistakes

Swapping the order in the numerator. $\dfrac{fg' - f'g}{g^2}$ gives the negative of the right answer. Start with the derivative of the top: “lo d-hi” comes first.

Forgetting to square the denominator, or forgetting the denominator entirely. Write the fraction bar and $\big[g(x)\big]^2$ before you fill in the numerator.

Dropping brackets in the numerator. In Example 1, $-(2x + 1)(1)$ is $-2x - 1$, not $-2x + 1$.

Dividing the derivatives. $\left(\dfrac{f}{g}\right)' \ne \dfrac{f'}{g'}$. Just like products, quotients need their own rule.

Mixing up the trig signs. The derivatives of $\cos x$, $\cot x$, and $\csc x$ are negative. Also, $\dfrac{d}{dx}[\sec x]$ is $\sec x \tan x$, not $\sec^2 x$ (that’s $\tan x$).

Expanding the denominator. Leave $(x^2 + 1)^2$ factored. Expanding it creates work and hides the fact that it’s never $0$.

Practice

1. (Warm-up) Differentiate $y = \dfrac{x}{x + 1}$.

Solution

$$\frac{dy}{dx} = \frac{(1)(x + 1) - (x)(1)}{(x + 1)^2} = \frac{1}{(x + 1)^2}$$

2. (Warm-up) Find $\dfrac{d}{dx}\big[\sec x + \tan x\big]$.

Solution

$$\sec x \tan x + \sec^2 x$$

3. (Warm-up) Let $q(x) = \dfrac{f(x)}{g(x)}$, where $f(1) = 6$, $f'(1) = 2$, $g(1) = 3$, and $g'(1) = -1$. Find $q'(1)$.

Solution

$$q'(1) = \frac{(2)(3) - (6)(-1)}{3^2} = \frac{6 + 6}{9} = \frac{4}{3}$$

4. (Core) Use the quotient rule to show that $\dfrac{d}{dx}[\cot x] = -\csc^2 x$.

Solution

Write $\cot x = \dfrac{\cos x}{\sin x}$:

$$\begin{aligned} \frac{d}{dx}[\cot x] &= \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} \\ &= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} \\ &= -\frac{1}{\sin^2 x} = -\csc^2 x \end{aligned}$$

5. (Core) Use the quotient rule to show that $\dfrac{d}{dx}[\csc x] = -\csc x \cot x$.

Solution

Write $\csc x = \dfrac{1}{\sin x}$:

$$\frac{d}{dx}[\csc x] = \frac{(0)(\sin x) - (1)(\cos x)}{\sin^2 x} = -\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = -\csc x \cot x$$

6. (Core) Find $\dfrac{dy}{dx}$ for $y = \dfrac{e^x}{x^2}$, and simplify.

Solution

$$\frac{dy}{dx} = \frac{e^x \cdot x^2 - e^x \cdot 2x}{x^4} = \frac{xe^x(x - 2)}{x^4} = \frac{e^x(x - 2)}{x^3}$$

7. (Core) Find the equation of the tangent line to $y = \dfrac{x^2 - 1}{x^2 + 1}$ at $x = 1$.

Solution

$$\frac{dy}{dx} = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2} = \frac{2x^3 + 2x - 2x^3 + 2x}{(x^2 + 1)^2} = \frac{4x}{(x^2 + 1)^2}$$

At $x = 1$: the point is $(1, 0)$ and the slope is $\dfrac{4}{4} = 1$.

$$y = x - 1$$

8. (Challenge) Show that the derivative of $y = \dfrac{\sin x}{1 + \cos x}$ simplifies to $\dfrac{1}{1 + \cos x}$.

Solution

$$\begin{aligned} \frac{dy}{dx} &= \frac{(\cos x)(1 + \cos x) - (\sin x)(-\sin x)}{(1 + \cos x)^2} \\ &= \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2} \\ &= \frac{\cos x + 1}{(1 + \cos x)^2} \\ &= \frac{1}{1 + \cos x} \end{aligned}$$

9. (Challenge) Find the point where the graph of $y = \dfrac{\ln x}{x}$ has a horizontal tangent.

Solution

$$\frac{dy}{dx} = \frac{\frac{1}{x}\cdot x - (\ln x)(1)}{x^2} = \frac{1 - \ln x}{x^2}$$

For $x \gt 0$ the denominator is positive, so set the numerator to $0$: $\ln x = 1$, so $x = e$. Then $y = \dfrac{\ln e}{e} = \dfrac{1}{e}$.

The point is $\left(e, \dfrac{1}{e}\right)$.