Derivatives of tan & sec — Graphical and Algebraic

Have you ever wondered why some functions grow so much faster than others? Today we’re going to figure out the derivatives of tangent and secant – two trig functions that can shoot off to infinity in the blink of an eye. The coolest part? We’ll prove the same result three completely different ways: with algebra, with a picture on the unit circle, and with complex exponentials. By the end, you’ll see how all three roads lead to the same beautiful answer.

TipWhy Derivatives of tan and sec Matter

The tangent and secant functions show up whenever you deal with slopes and stretching:

• Navigation: GPS systems use tangent functions to convert between angles and distances on a map — their derivatives help correct course in real time
• Architecture: the slope of a roof is a tangent ratio — the derivative tells you how sensitive the slope is to changes in angle
• Optics: lenses bend light at angles described by tangent — understanding how fast that angle changes is key to designing cameras and telescopes
• Roller coasters: at steep sections, the track angle is large and tangent grows fast — its derivative (sec^2) tells engineers how rapidly the slope is increasing

Once you know the derivatives of sin and cos, the derivatives of tan and sec follow — and today we’ll prove them three different ways!

Topics Covered

• Derivative of \(\tan\theta = \sec^2\theta\) via the quotient rule
• Graphical differentiation of trig functions on the unit circle (arc \(\approx\) chord for small angles)
• Derivative of \(\sec\theta = \sec\theta\cdot\tan\theta\) via similar triangles
• Complex exponential approach: writing \(\tan\theta\) in terms of \(e^{i\theta}\)
• Long division and simplification of complex exponential expressions
• Power expansion for differentials: plugging in \(\theta + d\theta\) and keeping first-order infinitesimals

Lecture Video

Key Frames from the Lecture

What You Need to Know First

NoteWhat are tangent and secant?

Tangent is the ratio of sine to cosine:

\[\tan\theta = \frac{\sin\theta}{\cos\theta}\]

Secant is the reciprocal of cosine:

\[\sec\theta = \frac{1}{\cos\theta}\]

On the unit circle, \(\tan\theta\) is the length of the vertical segment from the x-axis to where the radius line hits the vertical tangent line at \((1, 0)\). Secant is the distance from the origin to that same intersection point.

NoteWhat is the quotient rule?

The quotient rule tells you how to differentiate a fraction of two functions:

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

A handy mnemonic: “low d-high minus high d-low, over the square of what’s below.”

We already know \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\), so we have everything we need to differentiate \(\frac{\sin x}{\cos x}\).

NoteWhat are complex exponentials?

Using Euler’s formula \(e^{i\theta} = \cos\theta + i\sin\theta\), we can write sine and cosine as:

\[\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}, \qquad \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\]

This means we can also write tangent as a ratio of exponentials:

\[\tan\theta = \frac{e^{i\theta} - e^{-i\theta}}{i(e^{i\theta} + e^{-i\theta})}\]

Complex exponentials are powerful because the derivative of \(e^{i\theta}\) is simply \(ie^{i\theta}\) — no sign-flipping to remember!

NoteWhat does “arc equals chord” mean for small angles?

When a point moves along the unit circle by a tiny angle \(d\theta\), it traces out a small arc of length \(d\theta\).

For very small angles, that curved arc is practically indistinguishable from the straight-line chord connecting the two endpoints. This is the key geometric idea:

\[\text{arc} \approx \text{chord} \quad \text{when } d\theta \text{ is small}\]

This approximation is what makes geometric proofs of trig derivatives work.

NoteWhat are similar triangles?

Two triangles are similar if they have the same angles (same shape, possibly different size). When triangles are similar, the ratios of corresponding sides are equal:

\[\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\]

In this lesson, we use similar triangles that appear on the unit circle when we nudge an angle by \(d\theta\). The tiny triangle formed by the infinitesimal change is similar to the big triangle formed by the secant and tangent lines.

Key Concepts

Derivative of \(\tan\theta\) via the Quotient Rule

Since \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), we apply the quotient rule with \(f(\theta) = \sin\theta\) and \(g(\theta) = \cos\theta\):

\[\frac{d}{d\theta}(\tan\theta) = \frac{\cos\theta \cdot \cos\theta - \sin\theta \cdot (-\sin\theta)}{\cos^2\theta}\]

\[= \frac{\cos^2\theta + \sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}\]

ImportantKey Idea: Derivative of Tangent

The derivative of tangent is secant squared. This follows from the quotient rule on sin/cos, with the Pythagorean identity doing all the heavy lifting to simplify the numerator to 1.

\[\boxed{\frac{d}{d\theta}(\tan\theta) = \sec^2\theta}\]

Explore – see \(\tan\theta\) and its derivative \(\sec^2\theta\):

Drag the slider for \(a\). The green dashed line is tangent to \(\tan x\) (blue), and its slope always equals \(\sec^2(a)\) (red dashed curve). Notice how the slope is always at least 1 — \(\tan x\) is always getting steeper!

Graphical Derivation on the Unit Circle

Instead of algebra, we can derive \(\frac{d}{d\theta}(\tan\theta) = \sec^2\theta\) using a picture.

On the unit circle, draw the radius at angle \(\theta\) and extend it until it hits the vertical tangent line at \(x = 1\). The vertical intercept on that line has height \(\tan\theta\), and the distance from the origin to the intercept point is \(\sec\theta\).

Now nudge \(\theta\) by a tiny amount \(d\theta\):

1. The radius rotates by \(d\theta\), sweeping the intercept point upward along the tangent line
2. The tiny triangle formed at the intercept point is similar to the big triangle with sides \(1\), \(\tan\theta\), and \(\sec\theta\)
3. The small displacement along the tangent line is \(\sec^2\theta \cdot d\theta\)
4. This displacement equals \(d(\tan\theta)\)

So \(\frac{d(\tan\theta)}{d\theta} = \sec^2\theta\) — the same answer, straight from the geometry!

Explore – see the unit circle construction for \(\tan\theta\) and \(\sec\theta\):

Drag \(\theta\) and watch the tangent height (red segment) and the secant length (purple label) change. Notice how fast \(\tan\theta\) grows as \(\theta\) approaches \(\pm\frac{\pi}{2}\)!

Derivative of \(\sec\theta\) via Similar Triangles

We want \(\frac{d}{d\theta}(\sec\theta)\). Using the same unit circle picture:

\(\sec\theta\) is the distance from the origin to the point \((1, \tan\theta)\) on the vertical tangent line.

When \(\theta\) increases by \(d\theta\), the intercept point moves along the tangent line. The tiny triangle formed is similar to the original right triangle with legs \(1\) and \(\tan\theta\) and hypotenuse \(\sec\theta\).

By similar triangles, the change in the hypotenuse (which is \(d(\sec\theta)\)) satisfies:

\[\frac{d(\sec\theta)}{\sec\theta \cdot d\theta} = \frac{\sec\theta}{1} \cdot \frac{\tan\theta}{\sec\theta}\]

Working this out:

\[d(\sec\theta) = \sec\theta \cdot \tan\theta \cdot d\theta\]

ImportantKey Idea: Derivative of Secant

The derivative of secant is secant times tangent. You can prove this geometrically using similar triangles on the unit circle, or algebraically using the chain rule on \((\cos\theta)^{-1}\).

\[\boxed{\frac{d}{d\theta}(\sec\theta) = \sec\theta \cdot \tan\theta}\]

TipA quick algebraic check

We can verify this using the chain rule on \(\sec\theta = (\cos\theta)^{-1}\):

\[\frac{d}{d\theta}(\cos\theta)^{-1} = -1 \cdot (\cos\theta)^{-2} \cdot (-\sin\theta) = \frac{\sin\theta}{\cos^2\theta} = \frac{1}{\cos\theta} \cdot \frac{\sin\theta}{\cos\theta} = \sec\theta \cdot \tan\theta \;\checkmark\]

Complex Exponential Approach to \(\frac{d}{d\theta}(\tan\theta)\)

Here’s a slick algebraic method. Write tangent using Euler’s formula:

\[\tan\theta = \frac{e^{i\theta} - e^{-i\theta}}{i(e^{i\theta} + e^{-i\theta})}\]

To differentiate, we can use the quotient rule on this expression. Let \(u = e^{i\theta} - e^{-i\theta}\) and \(v = i(e^{i\theta} + e^{-i\theta})\). Then:

\[u' = ie^{i\theta} + ie^{-i\theta} = i(e^{i\theta} + e^{-i\theta})\]

\[v' = i(ie^{i\theta} - ie^{-i\theta}) = i^2(e^{i\theta} - e^{-i\theta}) = -(e^{i\theta} - e^{-i\theta})\]

By the quotient rule:

\[\frac{d}{d\theta}(\tan\theta) = \frac{u'v - uv'}{v^2}\]

\[= \frac{i(e^{i\theta}+e^{-i\theta}) \cdot i(e^{i\theta}+e^{-i\theta}) - (e^{i\theta}-e^{-i\theta}) \cdot [-(e^{i\theta}-e^{-i\theta})]}{[i(e^{i\theta}+e^{-i\theta})]^2}\]

The numerator becomes:

\[i^2(e^{i\theta}+e^{-i\theta})^2 + (e^{i\theta}-e^{-i\theta})^2\]

\[= -(e^{i\theta}+e^{-i\theta})^2 + (e^{i\theta}-e^{-i\theta})^2\]

Expanding both squares:

\[= -(e^{2i\theta} + 2 + e^{-2i\theta}) + (e^{2i\theta} - 2 + e^{-2i\theta}) = -4\]

The denominator is:

\[-\left(e^{i\theta}+e^{-i\theta}\right)^2 = -(2\cos\theta)^2 = -4\cos^2\theta\]

So:

\[\frac{d}{d\theta}(\tan\theta) = \frac{-4}{-4\cos^2\theta} = \frac{1}{\cos^2\theta} = \sec^2\theta \;\checkmark\]

Three completely different methods — quotient rule, geometry, and complex exponentials — all give \(\sec^2\theta\).

Long Division with Complex Exponentials

When working with expressions like \(\frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}}\), you can simplify by dividing numerator and denominator by \(e^{-i\theta}\):

\[\frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = \frac{e^{2i\theta} - 1}{e^{2i\theta} + 1}\]

You can also let \(w = e^{i\theta}\) to turn everything into a clean polynomial-style fraction:

\[\tan\theta = \frac{w^2 - 1}{i(w^2 + 1)} \quad \text{where } w = e^{i\theta}\]

This substitution makes the quotient rule much cleaner and reduces the chance of errors. The key trick is choosing a good substitution to simplify the algebra before you differentiate.

Power Expansion for Differentials

Another approach is to plug \(\theta + d\theta\) directly into the function and keep only first-order terms (dropping anything with \((d\theta)^2\) or higher):

For \(\tan(\theta + d\theta)\), recall that:

\[e^{i(\theta+d\theta)} = e^{i\theta} \cdot e^{i\,d\theta} \approx e^{i\theta}(1 + i\,d\theta)\]

since \(e^{i\,d\theta} \approx 1 + i\,d\theta\) when \(d\theta\) is infinitesimally small. Substituting into the tangent formula:

\[\tan(\theta + d\theta) \approx \frac{e^{i\theta}(1+i\,d\theta) - e^{-i\theta}(1-i\,d\theta)}{i\left[e^{i\theta}(1+i\,d\theta) + e^{-i\theta}(1-i\,d\theta)\right]}\]

Expanding the numerator:

\[(e^{i\theta} - e^{-i\theta}) + i\,d\theta\,(e^{i\theta} + e^{-i\theta})\]

Expanding the denominator:

\[i\left[(e^{i\theta} + e^{-i\theta}) + i\,d\theta\,(e^{i\theta} - e^{-i\theta})\right]\]

After careful algebra (dividing and dropping second-order terms), the result is:

\[\tan(\theta + d\theta) - \tan\theta = \sec^2\theta \cdot d\theta\]

This “infinitesimal substitution” method is very hands-on: you see exactly where each piece of the derivative comes from.

TipWhy can we drop \((d\theta)^2\)?

An infinitesimal \(d\theta\) is thought of as “so small that its square is essentially zero.” More precisely, when we take the limit as \(d\theta \to 0\) in \(\frac{\Delta f}{d\theta}\), any term with \((d\theta)^2\) in the numerator contributes \(0\) after dividing by \(d\theta\). So we only need to track first-order terms — everything else vanishes.

Explore – compare \(\tan(x)\) with its linear approximation at a point:

Drag \(a\) and see how the orange tangent line (slope \(= \sec^2 a\)) hugs the blue curve near the point. The green dashed curves are \(\pm\sec x\) — they form the “envelope” that \(\tan x\) bounces between.

Cheat Sheet

Formula	How to remember it
\(\dfrac{d}{d\theta}(\tan\theta) = \sec^2\theta\)	Quotient rule on \(\frac{\sin}{\cos}\), then Pythagorean identity
\(\dfrac{d}{d\theta}(\sec\theta) = \sec\theta\cdot\tan\theta\)	Chain rule on \((\cos\theta)^{-1}\), or similar triangles
\(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\)	SOH/CAH: opposite over adjacent
\(\sec\theta = \dfrac{1}{\cos\theta}\)	“Secant” = reciprocal of cosine
\(\sin^2\theta + \cos^2\theta = 1\)	The Pythagorean identity — the engine behind \(\sec^2\theta\)
\(1 + \tan^2\theta = \sec^2\theta\)	Divide the Pythagorean identity by \(\cos^2\theta\)

Complex Exponential Forms

\[\tan\theta = \frac{e^{i\theta} - e^{-i\theta}}{i(e^{i\theta} + e^{-i\theta})}\]

All Six Trig Derivatives (So Far)

\[\frac{d}{d\theta}(\sin\theta) = \cos\theta \qquad \frac{d}{d\theta}(\cos\theta) = -\sin\theta\]

\[\frac{d}{d\theta}(\tan\theta) = \sec^2\theta \qquad \frac{d}{d\theta}(\sec\theta) = \sec\theta\tan\theta\]

The Infinitesimal Trick

To find the derivative of \(f(\theta)\):

1. Compute \(f(\theta + d\theta)\) using \(e^{i\,d\theta} \approx 1 + i\,d\theta\)
2. Subtract \(f(\theta)\)
3. Keep only terms with one factor of \(d\theta\)
4. Divide by \(d\theta\) — that’s your derivative!