Math · Calculus I · Worked example

Differentiate tan x with the quotient rule

Use the quotient rule to show that the derivative of $tan x$ is ${sec}^{2} x$ , and say where the formula holds.

f (x) = tan x

Write the tangent as a quotient

The tangent is the sine divided by the cosine. Written that way it has a top and a bottom, and the quotient rule applies wherever $cos x \neq 0$ .

f (x) = \frac{sin x}{cos x}

Name the four pieces

The top is $sin x$ , with derivative $cos x$ . The bottom is $cos x$ , with derivative $- sin x$ . Both work as written only with $x$ in radians.

top = sin x, {top}^{'} = cos x, bottom = cos x, {bottom}^{'} = - sin x

\begin{array}{c} top = sin x, {top}^{'} = cos x \\ bottom = cos x \\ {bottom}^{'} = - sin x \end{array}

Apply the quotient rule

Bottom times the top’s derivative, minus top times the bottom’s derivative, over the bottom squared. Keep $- sin x$ in brackets for now.

f^{'} (x) = \frac{cos x cos x - sin x (- sin x)}{{cos}^{2} x}

\begin{aligned} f^{'} (x) \\ = \frac{cos x cos x - sin x (- sin x)}{{cos}^{2} x} \end{aligned}

Simplify the numerator

Subtracting $- {sin}^{2} x$ adds ${sin}^{2} x$ . The numerator becomes ${cos}^{2} x + {sin}^{2} x$ .

f^{'} (x) = \frac{{cos}^{2} x + {sin}^{2} x}{{cos}^{2} x}

Use the Pythagorean identity

For every angle, ${sin}^{2} x + {cos}^{2} x = 1$ , which makes the numerator $1$ .

f^{'} (x) = \frac{1}{{cos}^{2} x}

Name the result

Since $1 / cos x$ is $sec x$ , $1 / {cos}^{2} x$ is ${sec}^{2} x$ . The formula holds wherever $cos x \neq 0$ . That excludes $x = π / 2 + k π$ for every integer $k$ , the same places where $tan x$ is undefined.

f^{'} (x) = {sec}^{2} x

Read the answer

A square is never negative, and ${cos}^{2} x$ is at most $1$ , so ${sec}^{2} x$ is at least $1$ . The tangent rises on every interval between its asymptotes. At $x = 0$ its slope is exactly $1$ , and at $x = π / 4$ it is $1 / (1 / 2) = 2$ .

Result

The derivative of $tan x$ is ${sec}^{2} x$ , for every $x$ with $cos x \neq 0$ .

\frac{d}{d x} tan x = {sec}^{2} x

Your turn

1. One like the example

Differentiate $y = cot x$ by writing it as $cos x / sin x$ .

Show the answer and explanation

$d y / d x = - {csc}^{2} x$ , wherever $sin x \neq 0$ .

The top $cos x$ has derivative $- sin x$ and the bottom $sin x$ has derivative $cos x$ . The rule gives $(- sin x \cdot sin x - cos x \cdot cos x) / {sin}^{2} x$ $= - ({sin}^{2} x + {cos}^{2} x) / {sin}^{2} x$ $= - 1 / {sin}^{2} x$ , which is $- {csc}^{2} x$ .

\begin{array}{l} y = \frac{cos x}{sin x} \\ \frac{d y}{d x} = \frac{- sin x sin x - cos x cos x}{{sin}^{2} x} \\ \frac{d y}{d x} = - \frac{1}{{sin}^{2} x} \\ \frac{d y}{d x} = - {csc}^{2} x \end{array}

2. A variation

Differentiate $y = sin x / (1 + cos x)$ , and simplify.

Show the answer and explanation

$d y / d x = 1 / (1 + cos x)$ , wherever $cos x \neq - 1$ .

The rule gives $[cos x (1 + cos x) - sin x (- sin x)] / {(1 + cos x)}^{2}$ . The numerator is $cos x + {cos}^{2} x + {sin}^{2} x$ $= cos x + 1$ . One factor of $1 + cos x$ then cancels with the bottom.

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

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

3. Somewhere new

Find the equation of the tangent line to $y = tan x$ at $x = π / 4$ .

Show the answer and explanation

$y = 1 + 2 (x - π / 4)$ , a line of slope $2$ through $(π / 4, 1)$ .

The point is $(π / 4, 1)$ , because $tan (π / 4) = 1$ . The slope is ${sec}^{2} (π / 4)$ $= 1 / {cos}^{2} (π / 4)$ $= 1 / (1 / 2)$ $= 2$ . Point-slope form gives $y = 1 + 2 (x - π / 4)$ .

\begin{array}{l} f (x) = tan x \\ f^{'} (x) = {sec}^{2} x \\ f^{'} (\frac{π}{4}) = 2 \end{array}

Keep exploring

Open the steps in Math: the Checker marks every row ✓. Swap the two terms of the numerator and that row reads ✗, because the order in the quotient rule matters.

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Write the tangent as a quotient

Name the four pieces

Apply the quotient rule

Simplify the numerator

Use the Pythagorean identity

Name the result

Read the answer

Result

Your turn

1. One like the example

d⁢y⁢/⁢d⁢x=−csc2xdy/dx=−csc2x, wherever sinx≠0sinx=0.

2. A variation

d⁢y⁢/⁢d⁢x=1⁢/⁢(1+cosx)dy/dx=1/(1+cosx), wherever cosx≠−1cosx=−1.

3. Somewhere new

y=1+2⁢(x−π⁢/4)y=1+2(x−π/4), a line of slope 22 through (π⁢/4,1)(π/4,1).

Keep exploring

Study it, then try it

$d y / d x = - {csc}^{2} x$ , wherever $sin x \neq 0$ .

$d y / d x = 1 / (1 + cos x)$ , wherever $cos x \neq - 1$ .

$y = 1 + 2 (x - π / 4)$ , a line of slope $2$ through $(π / 4, 1)$ .