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Trigonometric Derivatives · Tutorial 491 of 1495

Deriving the Tangent Derivative by the Quotient Rule

Use the quotient rule and a trigonometric identity to derive the tangent derivative, understand its domain, and check the result in equivalent forms.

Intermediate 9 min read

What You'll Learn

  • Derive the derivative of tangent from its sine-over-cosine definition.
  • Identify where tangent and its derivative are defined.
  • Rewrite the result using secant squared and one plus tangent squared.
  • Evaluate the derivative at exact angles and verify each value in more than one form.
  • Recognize and correct a common quotient-rule sign error.

Why Derive the Tangent Derivative?

Tangent is defined as a quotient of two familiar trigonometric functions. That makes its derivative a direct application of the quotient rule, but the signs matter: the derivative of cosine is negative sine, and the quotient rule subtracts the denominator derivative term. Keeping both facts visible explains why the result is positive and how it can be written in several equivalent ways.

Throughout, angles are measured in radians. The earlier results for sine and cosine are \((\sin x)'=\cos x\) and \((\cos x)'=-\sin x\). We will use these derivatives, the quotient rule, and the identity \(\sin^2 x+\cos^2 x=1\) to derive the formula.

Definition and Domain

Definition: The tangent function is defined by \(\tan x=\dfrac{\sin x}{\cos x}\) wherever \(\cos x\ne 0\). Thus its domain is the set of real numbers \(x\) for which cosine is nonzero.

At an angle where \(\cos x=0\), the quotient defining tangent has a zero denominator. Tangent is therefore undefined there, so it cannot have a derivative there. Equivalently, its domain is \(\mathbb{R}\) with the angles \(\dfrac{\pi}{2}+k\pi\), for \(k\in\mathbb{Z}\), excluded.

The restriction is also needed when applying the quotient rule: the denominator function must be nonzero at the point under consideration. The resulting derivative formula will consequently be valid at every point in the domain of tangent, and nowhere outside that domain.

Deriving the Formula

For differentiable functions \(u\) and \(v\), with \(v(x)\ne 0\), the quotient rule states that

$$ \left(\frac{u}{v}\right)'(x) = \frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}. $$

For tangent, take \(u(x)=\sin x\) and \(v(x)=\cos x\). Their derivatives are \(u'(x)=\cos x\) and \(v'(x)=-\sin x\). Substitution gives

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

The first minus sign in the quotient rule combines with the negative sign in the derivative of cosine. Their product contributes a positive \(\sin^2 x\) term. The last equality uses the identity \(\sin^2 x+\cos^2 x=1\). Since \(\cos x\ne0\) on the domain of tangent, every denominator in this calculation is nonzero there.

Theorem: For every real \(x\) such that \(\cos x\ne0\), \(\dfrac{d}{dx}(\tan x)=\dfrac{1}{\cos^2 x}=\sec^2 x=1+\tan^2 x\).

The first equivalent form is \(\sec^2 x\), since \(\sec x\) is defined as \(1/\cos x\) wherever cosine is nonzero. For the other form, divide \(\sin^2 x+\cos^2 x=1\) by \(\cos^2 x\), which is allowed on the same domain:

$$ \frac{\sin^2 x}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} \quad\Longrightarrow\quad \tan^2 x+1=\sec^2 x. $$

Thus the expressions \(\dfrac{1}{\cos^2 x}\), \(\sec^2 x\), and \(1+\tan^2 x\) give the same derivative at every point where tangent is defined. These are alternative forms of one result, not formulas with different domains.

Worked Examples

Worked Example: Find the Derivative at \(x=\dfrac{\pi}{4}\)

First check the domain: \(\cos(\pi/4)=\sqrt{2}/2\ne0\), so the derivative exists. Using the reciprocal-cosine form,

$$ \sec^2\left(\frac{\pi}{4}\right) = \frac{1}{\cos^2(\pi/4)} = \frac{1}{(\sqrt{2}/2)^2} = \frac{1}{2/4} = 2. $$

Check with the tangent form. Since \(\tan(\pi/4)=1\),

$$ 1+\tan^2\left(\frac{\pi}{4}\right) = 1+1^2 = 2. $$

The two forms agree, so the derivative value is \(2\).

Worked Example: Find the Derivative at \(x=\dfrac{\pi}{6}\) Directly from the Quotient

At \(x=\pi/6\), the exact trigonometric values are \(\sin(\pi/6)=1/2\) and \(\cos(\pi/6)=\sqrt{3}/2\). Applying the quotient-rule expression before simplifying gives

$$ \begin{aligned} \frac{d}{dx}(\tan x)\bigg|_{x=\pi/6} &= \frac{\cos^2(\pi/6)+\sin^2(\pi/6)}{\cos^2(\pi/6)}\\ &= \frac{(\sqrt{3}/2)^2+(1/2)^2}{(\sqrt{3}/2)^2}\\ &= \frac{3/4+1/4}{3/4}\\ &= \frac{1}{3/4} = \frac{4}{3}. \end{aligned} $$

As a check, \(\tan(\pi/6)=1/\sqrt{3}\), so the other equivalent form gives

$$ 1+\tan^2\left(\frac{\pi}{6}\right) = 1+\left(\frac{1}{\sqrt{3}}\right)^2 = 1+\frac{1}{3} = \frac{4}{3}. $$

Both calculations give the same derivative value.

Worked Example: Check a Negative Angle

Find the derivative at \(x=-\pi/4\). Here \(\cos(-\pi/4)=\sqrt{2}/2\), which is nonzero, so the formula applies. Then

$$ \sec^2\left(-\frac{\pi}{4}\right) = \frac{1}{(\sqrt{2}/2)^2} = 2. $$

Also, \(\tan(-\pi/4)=-1\), and therefore

$$ 1+\tan^2\left(-\frac{\pi}{4}\right) = 1+(-1)^2 = 2. $$

The negative value of tangent is squared in this form; it does not make the derivative negative. The two equivalent expressions verify the derivative value \(2\).

Worked Example: Identify an Angle Where the Formula Does Not Apply

Consider \(x=\pi/2\). Since \(\cos(\pi/2)=0\), the defining quotient for tangent is undefined:

$$ \tan\left(\frac{\pi}{2}\right) = \frac{\sin(\pi/2)}{\cos(\pi/2)} = \frac{1}{0}. $$

There is no tangent value at this angle, and hence no derivative of tangent there. The expression \(1/\cos^2(\pi/2)\) also has a zero denominator. This confirms that the domain condition is essential, rather than an optional restriction on the formula.

A Common Quotient-Rule Sign Error

A frequent mistake is to change the quotient rule's subtraction into addition, or to lose the negative sign in \((\cos x)'=-\sin x\). For example, incorrectly using \(u'v+uv'\) would give

$$ \frac{(\cos x)(\cos x)+(\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x-\sin^2 x}{\cos^2 x}. $$

At \(x=\pi/4\), this incorrect expression becomes

$$ \frac{1/2-1/2}{1/2}=0. $$

That is not the derivative. The quotient rule has a minus between its two numerator terms, so the correct numerator is \(\cos x\cos x-\sin x(-\sin x)\). At \(x=\pi/4\), the corrected calculation is

$$ \frac{1/2-(-1/2)}{1/2} = \frac{1}{1/2} = 2, $$

which agrees with \(\sec^2(\pi/4)=2\). Keeping the quotient-rule subtraction and the negative cosine derivative visible until they have both been substituted prevents this error.

How to Use the Result Reliably

When applying the tangent derivative formula, first check that the input lies in the domain. Then choose whichever equivalent form makes the arithmetic clearest. The following sequence records those checks:

1
Check the angle.
Verify that \(\cos x\ne0\); otherwise tangent itself is undefined.
2
Select a form.
Use \(1/\cos^2 x\), \(\sec^2 x\), or \(1+\tan^2 x\), according to which values are known.
3
Evaluate and verify.
Substitute exact trigonometric values, simplify, and check the result with another equivalent form when useful.

The domain check matters even if a symbolic expression looks simple. For example, the identity \(1+\tan^2 x=\sec^2 x\) does not assign a value at an angle where tangent and secant are both undefined. An identity can be used only where its expressions are defined.

Why the Equivalent Forms Matter

The quotient-rule form makes the derivation transparent: it shows exactly how the sine and cosine derivatives combine. The form \(\sec^2 x\) is compact, while \(1+\tan^2 x\) can be useful when a tangent value is already known. Because these forms agree on the domain, they provide independent checks against arithmetic and sign errors.

The derivation also illustrates a general strategy in differentiation: represent a function using operations whose derivative rules are known, apply the appropriate rule with its conditions, and simplify using an identity only where the expressions are defined. Here, writing tangent as a quotient turns its derivative into a consequence of the sine and cosine derivatives.

Check Your Understanding

Answer each question, paying attention to both the calculation and the domain.

  1. Starting from \(\tan x=\sin x/\cos x\), what quotient-rule numerator do you obtain before simplifying?
  2. Why must \(\cos x\ne0\) when differentiating tangent by the quotient rule?
  3. Use \(1+\tan^2 x\) to find the derivative value at \(x=\pi/3\).
  4. Explain why the derivative value at \(x=-\pi/4\) is positive even though the tangent value is negative.
  5. What sign error produces \((\cos^2 x-\sin^2 x)/\cos^2 x\), and what is the correct numerator instead?

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