Why is the derivative of tan x equal to sec squared x?

Because tangent is \frac{\sin x}{\cos x}. When you apply the quotient rule, the numerator becomes \cos^2 x + \sin^2 x, which equals 1 by the Pythagorean identity. The denominator is \cos^2 x, so you get \frac{1}{\cos^2 x} = \sec^2 x.

How do I use the derivative of tangent in a problem?

Use it whenever you need the slope of y = \tan x at a point, or when you are differentiating a more complex function containing tangent. For composite functions like \tan(3x) or 5\tan(x^2), combine this derivative with the chain rule.

Is the derivative of tan x always positive?

Yes, because \sec^2 x = \frac{1}{\cos^2 x} is always positive (it is the reciprocal of a squared real number). This means \tan x is always increasing, even though the curve has vertical asymptotes where it jumps from -\infty to +\infty.

Can I use the derivative of tangent to solve optimization problems?

Yes. To find critical points where y = \tan x has maximum or minimum slope, or to solve applied problems where a tangent function describes a quantity (like an angle or position), set \frac{dy}{dx} = \sec^2 x equal to values from your problem constraints.

Derivative of Tan x

The derivative of tan x equals sec squared x, needed to find slopes of tangent curves and solve optimization and related-rates problems.

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

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What each symbol means

What Derivative of Tan x takes

Symbol	Meaning
$x$	The independent variable, typically representing an angle in radians; if you accidentally use degrees instead, your derivative will be off by a constant factor.

When to use it

Use this when you need the slope of $y = an x$ at any point, or when differentiating a function that contains tangent.

Level

Usually taught in: Calculus I

Worked examples

1. Find the derivative of $y = \tan x$

Problem

Find the derivative of

y = \tan x

with respect to

x

$\frac{d}{dx}\tan x = \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)$
We rewrite tangent as a quotient so we can apply the quotient rule.
$= \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x}$
The quotient rule gives us $\frac{u'v - uv'}{v^2}$ . Here $u = \sin x$ with $u' = \cos x$ , and $v = \cos x$ with $v' = -\sin x$ .
$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$
We distribute and simplify: $-(-\sin x) = +\sin x$ .
$= \frac{1}{\cos^2 x}$
The Pythagorean identity states that $\cos^2 x + \sin^2 x = 1$ .
$= \sec^2 x$
By definition, $\sec x = \frac{1}{\cos x}$ , so $\sec^2 x = \frac{1}{\cos^2 x}$ .

Answer: $\frac{dy}{dx} = \sec^2 x$

This derivation shows why the formula works: tangent is a ratio, and the quotient rule combined with the Pythagorean identity gives us this elegant result. Understanding the derivation helps you remember the formula and recognize when to use it.

2. Find the derivative of $y = 5\tan(3x)$ using the chain rule

Problem

Find the derivative of

y = 5\tan(3x)

with respect to

x

$\frac{dy}{dx} = 5 \cdot \frac{d}{dx}\tan(3x)$
The constant 5 comes out front, and we need to differentiate $\tan(3x)$ .
$= 5 \cdot \sec^2(3x) \cdot \frac{d}{dx}(3x)$
Apply the chain rule: differentiate the outside function (tangent gives $\sec^2$ ) and multiply by the derivative of the inside function $3x$ .
$= 5 \cdot \sec^2(3x) \cdot 3$
The derivative of $3x$ is 3.
$= 15\sec^2(3x)$
Multiply the constants: $5 \times 3 = 15$ .

Answer: $\frac{dy}{dx} = 15\sec^2(3x)$

Whenever tangent's input is not just $x$ , the chain rule is essential. This example shows how to combine the derivative formula with the chain rule to handle more complex functions.

3. Rate of change of a rotating thermometer pointer

Problem

In a laboratory, a thermometer with a rotating pointer displays temperature readings. The pointer's angle (in radians) at time

t

seconds is

\theta(t) = \tan(0.2t)

. Find the rate of change of the pointer angle at

t = 3

seconds.

$\frac{d\theta}{dt} = \frac{d}{dt}\tan(0.2t)$
We need the derivative of the angle with respect to time.
$= \sec^2(0.2t) \cdot \frac{d}{dt}(0.2t)$
Apply the formula $\frac{d}{dx}\tan x = \sec^2 x$ along with the chain rule for the inner function.
$= \sec^2(0.2t) \cdot 0.2$
The derivative of $0.2t$ is $0.2$ .
$\frac{d\theta}{dt}\bigg|_{t=3} = 0.2 \cdot \sec^2(0.2 \times 3)$
Substitute $t = 3$ seconds to find the instantaneous rate at this specific time.
$= 0.2 \cdot \sec^2(0.6)$
Compute the argument: $0.2 \times 3 = 0.6$ radians.
$\approx 0.2 \times (1.086)^2 \approx 0.236 \text{ rad/s}$
Using a calculator, $\sec(0.6) \approx 1.086$ , so $\sec^2(0.6) \approx 1.179$ , and $0.2 \times 1.179 \approx 0.236$ radians per second.

Answer: $\frac{d\theta}{dt}\bigg|_{t=3} \approx 0.236 \text{ rad/s}$

This word problem demonstrates that derivatives of tangent appear naturally when measuring how quickly an angle or oscillating quantity changes over time. By combining the derivative formula with the chain rule, we found the exact rate of change at a specific moment.

Common mistakes

Where Derivative of Tan x usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\frac{d}{dx}\tan x = \tan^2 x$	This confuses the tangent derivative with the power rule, as if you were differentiating $\tan x$ like $(\text{something})^2$ .	The correct formula is $\frac{d}{dx}\tan x = \sec^2 x$ (secant squared, not tangent squared). Remember that $\sec x = \frac{1}{\cos x}$ is completely different from $\tan x = \frac{\sin x}{\cos x}$ .
Writing $\frac{d}{dx}\tan(2x) = \sec^2(2x)$ and forgetting the chain rule	You apply the outer rule correctly but forget to multiply by the derivative of the inner function $2x$ .	Always include the chain rule factor: $\frac{d}{dx}\tan(2x) = \sec^2(2x) \cdot 2 = 2\sec^2(2x)$ .
Writing $\frac{d}{dx}\tan x = \sec x$ (missing the square)	You remember that tangent derivatives involve secant but forget that the exponent is 2, not 1.	The correct formula has a square: $\frac{d}{dx}\tan x = \sec^2 x$ , not $\sec x$ . A key to remembering: the quotient rule on $\frac{\sin x}{\cos x}$ produces $\sec^2 x$ in the final simplified form.

Tips and when to use something else

Always write $\sec^2 x$ , not $\sec x$ and not $\tan^2 x$ —the square is essential to the formula.
When you see $\tan(kx)$ for any constant $k$ , do not forget the chain rule: multiply your answer by $k$ .
If a problem asks about tangent but is not asking for a derivative, use a different method instead—for example, use the Product Rule if you're multiplying tangent by another function, or integrate if you need an antiderivative.
The derivative $\sec^2 x$ is always positive, so $\tan x$ is always increasing. This is why the graph of tangent gets steeper as it approaches its vertical asymptotes at odd multiples of $\pi/2$ .

Related formulas

Quotient RuleThe Quotient Rule gives you the derivative of a fraction by combining the derivatives and original functions of the numerator and denominator.Chain RuleThe Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.Derivative of Sin xThe derivative of sin x is cos x, giving the slope of the sine curve; use this rule when differentiating any sine function.Derivative of Cos xThe derivative of cosine x equals negative sine x, the fundamental rule for finding rates of change of oscillating cosine functions.Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.

Frequently asked questions

Why is the derivative of tan x equal to sec squared x?: Because tangent is $\frac{\sin x}{\cos x}$ . When you apply the quotient rule, the numerator becomes $\cos^2 x + \sin^2 x$ , which equals 1 by the Pythagorean identity. The denominator is $\cos^2 x$ , so you get $\frac{1}{\cos^2 x} = \sec^2 x$ .
How do I use the derivative of tangent in a problem?: Use it whenever you need the slope of $y = \tan x$ at a point, or when you are differentiating a more complex function containing tangent. For composite functions like $\tan(3x)$ or $5\tan(x^2)$ , combine this derivative with the chain rule.
Is the derivative of tan x always positive?: Yes, because $\sec^2 x = \frac{1}{\cos^2 x}$ is always positive (it is the reciprocal of a squared real number). This means $\tan x$ is always increasing, even though the curve has vertical asymptotes where it jumps from $-\infty$ to $+\infty$ .
Can I use the derivative of tangent to solve optimization problems?: Yes. To find critical points where $y = \tan x$ has maximum or minimum slope, or to solve applied problems where a tangent function describes a quantity (like an angle or position), set $\frac{dy}{dx} = \sec^2 x$ equal to values from your problem constraints.

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Reviewed 2026-09-18