Trig Derivatives: Tan, Cot, Sec & Csc | AP Calculus 2.10

1. Four essential trigonometric derivative formulas

Core concept: Complete the six basic trigonometric derivatives by deriving the formulas for $\tan x$ , $\cot x$ , $\sec x$ and $\csc x$ . Rewrite each function as a quotient of sine and cosine, then apply the quotient rule. Use radian measure and keep the original domain. The sine and cosine derivatives supply the starting formulas.

MEMORIZE THESE FOUR FORMULAS!

Function	Derivative	Leading sign in formula
$\tan x$	$\sec^{2} x$	No explicit minus sign
$\cot x$	$- \csc^{2} x$	Explicit minus sign
$\sec x$	$\sec x \tan x$	No explicit minus sign
$\csc x$	$- \csc x \cot x$	Explicit minus sign

Formula pattern: The cotangent and cosecant formulas have an explicit leading minus sign. This mnemonic does not mean the numerical value of the cosecant derivative is always negative.

Read the notation: $\sec^{2} x = (\sec x)^{2}$ and $\csc^{2} x = (\csc x)^{2}$ . These squares differ from the product forms $\sec x \tan x$ and $\csc x \cot x$ . All formulas use radians and apply only where the original function is defined.

2. Derive tangent, cotangent, secant and cosecant

DERIVATIVE 1: TANGENT FUNCTION

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

In Words: The derivative of tangent is secant squared.

Proof (Using Quotient Rule):

Rewrite using sin and cos:
$\tan x = \frac{\sin x}{\cos x}$
Apply Quotient Rule: "Low d High minus High d Low over Low Low"
$\frac{d}{d x} [\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (- \sin x)}{\cos^{2} x}$
Simplify (double negative becomes positive):
$= \frac{\cos^{2} x + \sin^{2} x}{\cos^{2} x}$
Use Pythagorean Identity $\sin^{2} x + \cos^{2} x = 1$ :
$= \frac{1}{\cos^{2} x} = \sec^{2} x ✓$

Alternative Identity Connection:

The identity $1 + \tan^{2} x = \sec^{2} x$ rewrites the derivative as $\frac{d}{d x} [\tan x] = 1 + \tan^{2} x$ . It is an equivalent form of the result, rather than a proof of the derivative by itself.

Examples:

$\frac{d}{d x} [\tan x] = \sec^{2} x$
$\frac{d}{d x} [5 \tan x] = 5 \sec^{2} x$
$\frac{d}{d x} [\tan x + x^{2}] = \sec^{2} x + 2 x$

DERIVATIVE 2: COTANGENT FUNCTION

\frac{d}{d x} [\cot x] = - \csc^{2} x

In Words: The derivative of cotangent is negative cosecant squared.

CRITICAL: Don't forget the negative sign! This is the #1 mistake with cotangent.

Proof (Using Quotient Rule):

Rewrite using sin and cos:
$\cot x = \frac{\cos x}{\sin x}$
Apply Quotient Rule:
$\frac{d}{d x} [\cot x] = \frac{\sin x \cdot (- \sin x) - \cos x \cdot \cos x}{\sin^{2} x}$
Simplify:
$= \frac{- \sin^{2} x - \cos^{2} x}{\sin^{2} x} = \frac{- (\sin^{2} x + \cos^{2} x)}{\sin^{2} x}$
Use Pythagorean Identity:
$= \frac{- 1}{\sin^{2} x} = - \csc^{2} x ✓$

Alternative Identity Connection:

The identity $1 + \cot^{2} x = \csc^{2} x$ rewrites the derivative as $\frac{d}{d x} [\cot x] = - (1 + \cot^{2} x)$ . Keep the leading minus sign on the entire quantity.

Examples:

$\frac{d}{d x} [\cot x] = - \csc^{2} x$
$\frac{d}{d x} [4 \cot x] = - 4 \csc^{2} x$
$\frac{d}{d x} [- \cot x] = - (- \csc^{2} x) = \csc^{2} x$ (double negative!)

DERIVATIVE 3: SECANT FUNCTION

\frac{d}{d x} [\sec x] = \sec x \tan x

In Words: The derivative of secant is secant times tangent.

Proof Method 1 (Using Quotient Rule):

Rewrite as reciprocal:
$\sec x = \frac{1}{\cos x}$
Apply Quotient Rule:
$\frac{d}{d x} [\sec x] = \frac{\cos x \cdot 0 - 1 \cdot (- \sin x)}{\cos^{2} x} = \frac{\sin x}{\cos^{2} x}$
Rewrite as product:
$= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x ✓$

Proof Method 2: chain-rule preview for Unit 3

Rewrite as power: $\sec x = (\cos x)^{- 1}$
Apply Chain Rule: $- 1 (\cos x)^{- 2} \cdot (- \sin x) = \frac{\sin x}{\cos^{2} x}$
Convert: $= \sec x \tan x$

Examples:

$\frac{d}{d x} [\sec x] = \sec x \tan x$
$\frac{d}{d x} [3 \sec x] = 3 \sec x \tan x$
$\frac{d}{d x} [\sec x - \cos x] = \sec x \tan x + \sin x$

DERIVATIVE 4: COSECANT FUNCTION

\frac{d}{d x} [\csc x] = - \csc x \cot x

In Words: The derivative of cosecant is negative cosecant times cotangent.

CRITICAL: Don't forget the negative sign!

Proof Method 1 (Using Quotient Rule):

Rewrite as reciprocal:
$\csc x = \frac{1}{\sin x}$
Apply Quotient Rule:
$\frac{d}{d x} [\csc x] = \frac{\sin x \cdot 0 - 1 \cdot \cos x}{\sin^{2} x} = \frac{- \cos x}{\sin^{2} x}$
Rewrite as product:
$= - \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = - \csc x \cot x ✓$

Proof Method 2: chain-rule preview for Unit 3

Rewrite as power: $\csc x = (\sin x)^{- 1}$
Apply Chain Rule: $- 1 (\sin x)^{- 2} \cdot \cos x = - \frac{\cos x}{\sin^{2} x}$
Convert: $= - \csc x \cot x$

Examples:

$\frac{d}{d x} [\csc x] = - \csc x \cot x$
$\frac{d}{d x} [2 \csc x] = - 2 \csc x \cot x$
$\frac{d}{d x} [- \csc x] = - (- \csc x \cot x) = \csc x \cot x$ (double negative!)

Why these derivations work: Each identity is used on its original domain, with a nonzero denominator. Apply the quotient rule using radian sine and cosine derivatives, then simplify with $\sin^{2} x + \cos^{2} x = 1$ . The chain-rule alternatives for the reciprocal functions are retained as a preview; the quotient proofs establish all four formulas without that later rule.

3. All six trigonometric derivatives

Function	Derivative	Key Note
$\sin x$	$\cos x$	No explicit leading minus sign
$\cos x$	$- \sin x$	Explicit leading minus sign
$\tan x$	$\sec^{2} x$	Squared function
$\cot x$	$- \csc^{2} x$	Explicit leading minus sign
$\sec x$	$\sec x \tan x$	Product form
$\csc x$	$- \csc x \cot x$	Explicit leading minus sign

4. Domains and the actual signs of derivatives

The derivative formulas have the same domains as these original basic functions. Let $k$ be any integer.

Functions	Required condition	Excluded inputs
$\tan x$ , $\sec x$	$\cos x \neq 0$	$x \neq \frac{π}{2} + k π$
$\cot x$ , $\csc x$	$\sin x \neq 0$	$x \neq k π$

Sine and cosine, and their derivatives, are defined for all real inputs. In a sum or product, use the intersection of the original factor domains. Simplification must not fill a gap in the original expression.

A minus sign in a formula is different from a negative value

On their respective domains, $\sec^{2} x > 0$ and $- \csc^{2} x < 0$ , so tangent has positive derivative and cotangent has negative derivative on each interval between their excluded inputs. Secant and cosecant behave differently:

(\sec x)^{'} = \sec x \tan x = \frac{\sin x}{\cos^{2} x} .

(\csc x)^{'} = - \csc x \cot x = - \frac{\cos x}{\sin^{2} x} .

The secant derivative has the sign of $\sin x$ ; the cosecant derivative has the opposite sign of $\cos x$ . Either can be positive, negative or zero where the original function is defined. For example,

(\sec x)^{'} |_{x = - \frac{π}{6}} = - \frac{2}{3}, (\csc x)^{'} |_{x = \frac{2 π}{3}} = \frac{2}{3} .

Also, $(\sec x)^{'} |_{x = 0} = 0$ and $(\csc x)^{'} |_{x = \frac{π}{2}} = 0$ . These examples show why a mnemonic about leading signs must not be treated as a claim about increasing or decreasing behavior everywhere.

5. Patterns and memory aids

PATTERN RECOGNITION

Pattern 1: Leading minus signs in the formulas

The standard formulas for $\cos x$ , $\cot x$ and $\csc x$ have an explicit leading minus sign. The value of a derivative can still be positive, negative or zero depending on the remaining factors and the input.

$\cos x \to - \sin x$ (negative)
$\cot x \to - \csc^{2} x$ (negative)
$\csc x \to - \csc x \cot x$ (negative)

The standard formulas for $\sin x$ , $\tan x$ and $\sec x$ have no explicit leading minus sign. This does not make all three derivatives positive everywhere: for instance, $\frac{d}{d x} [\sec x] = \frac{\sin x}{\cos^{2} x}$ has the sign of sine.

Pattern 2: Squared vs. Product

The quotient functions $\tan x$ and $\cot x$ have derivatives involving squared reciprocal functions:

$\tan x = \frac{\sin x}{\cos x} \to \sec^{2} x$ (reciprocal of cosine is secant)
$\cot x = \frac{\cos x}{\sin x} \to - \csc^{2} x$ (reciprocal of sine is cosecant)

The reciprocal functions $\sec x$ and $\csc x$ have derivatives in product form:

$\sec x = \frac{1}{\cos x} \to \sec x \tan x$
$\csc x = \frac{1}{\sin x} \to - \csc x \cot x$

Pattern 3: Function Appears in Own Derivative

For $\sec x$ and $\csc x$ , the original function appears as a factor in the derivative:

$\sec x \to \sec x \tan x$
$\csc x \to - \csc x \cot x$

Mnemonic Devices:

For Tangent & Cotangent:

"Tan goes to sec squared,
Cot goes negative, csc squared!"

For Secant & Cosecant:

"Sec times tan is what you get,
Csc cot, but don't forget: it's negative!"

Negative Sign Rule:

"C" words have minus signs!
(Cos, Cot, Csc)

6. Six worked examples with full steps

Example 1: Basic Tangent

Problem: Find $f^{'} (x)$ for $f (x) = 3 \tan x - 2 x^{2}$

Solution:

Differentiate term-by-term:
- $\frac{d}{d x} [3 \tan x] = 3 \sec^{2} x$
- $\frac{d}{d x} [- 2 x^{2}] = - 4 x$
Combine: $f^{'} (x) = 3 \sec^{2} x - 4 x$

Answer: $f^{'} (x) = 3 \sec^{2} x - 4 x$

Example 2: Cotangent with Sign Watch

Problem: Differentiate $g (x) = 5 \cot x + \sin x$

Solution:

First term: $\frac{d}{d x} [5 \cot x] = 5 (- \csc^{2} x) = - 5 \csc^{2} x$
Second term: $\frac{d}{d x} [\sin x] = \cos x$
Combine: $g^{'} (x) = - 5 \csc^{2} x + \cos x$

Answer: $g^{'} (x) = - 5 \csc^{2} x + \cos x$

Example 3: Secant Function

Problem: Find $\frac{d y}{d x}$ for $y = x^{3} - 4 \sec x$

Solution:

First term: $\frac{d}{d x} [x^{3}] = 3 x^{2}$
Second term: $\frac{d}{d x} [- 4 \sec x] = - 4 (\sec x \tan x) = - 4 \sec x \tan x$
Combine: $\frac{d y}{d x} = 3 x^{2} - 4 \sec x \tan x$

Answer: $\frac{d y}{d x} = 3 x^{2} - 4 \sec x \tan x$

Example 4: Cosecant with Double Negative

Problem: Differentiate $h (x) = - \csc x + e^{x}$

Solution:

First term (watch signs!):
- $\frac{d}{d x} [- \csc x] = - 1 \cdot \frac{d}{d x} [\csc x]$
- $= - 1 \cdot (- \csc x \cot x)$
- $= \csc x \cot x$ (double negative becomes positive!)
Second term: $\frac{d}{d x} [e^{x}] = e^{x}$
Combine: $h^{'} (x) = \csc x \cot x + e^{x}$

Answer: $h^{'} (x) = \csc x \cot x + e^{x}$

Example 5: All Four Together

Problem: Find $k^{'} (x)$ for $k (x) = \tan x - \cot x + \sec x - \csc x$

Solution:

Differentiate each term:
- $\frac{d}{d x} [\tan x] = \sec^{2} x$
- $\frac{d}{d x} [- \cot x] = - (- \csc^{2} x) = \csc^{2} x$
- $\frac{d}{d x} [\sec x] = \sec x \tan x$
- $\frac{d}{d x} [- \csc x] = - (- \csc x \cot x) = \csc x \cot x$
Combine all:
$k^{'} (x) = \sec^{2} x + \csc^{2} x + \sec x \tan x + \csc x \cot x$

Answer: $k^{'} (x) = \sec^{2} x + \csc^{2} x + \sec x \tan x + \csc x \cot x$

Example 6: Mixed with Polynomial

Problem: Differentiate $f (x) = 2 \tan x + 3 \cot x - x^{2} + 5$

Solution:

f^{'} (x) = 2 \sec^{2} x + 3 (- \csc^{2} x) - 2 x + 0

f^{'} (x) = 2 \sec^{2} x - 3 \csc^{2} x - 2 x

Answer: $f^{'} (x) = 2 \sec^{2} x - 3 \csc^{2} x - 2 x$

Keep the domains: Examples 1 and 3 require $\cos x \neq 0$ . Examples 2 and 4 require $\sin x \neq 0$ . Examples 5 and 6 require both $\sin x \neq 0$ and $\cos x \neq 0$ . In every case, the derivative is stated on that original domain and trigonometric inputs use radians.

7. Evaluate derivatives and write tangent lines

Exact derivative values at a common angle

At $a = \frac{π}{4}$ , all four functions are defined. Use $\sin a = \cos a = \frac{\sqrt{2}}{2}$ , $\tan a = \cot a = 1$ and $\sec a = \csc a = \sqrt{2}$ .

Function	Function value at $a$	Derivative value at $a$
$\tan x$	$1$	$\sec^{2} a = 2$
$\cot x$	$1$	$- \csc^{2} a = - 2$
$\sec x$	$\sqrt{2}$	$\sec a \tan a = \sqrt{2}$
$\csc x$	$\sqrt{2}$	$- \csc a \cot a = - \sqrt{2}$

A tangent line uses both value and slope

For $f (x) = \tan x$ at $a = \frac{π}{4}$ , the point is $(\frac{π}{4}, 1)$ and the slope is $f^{'} (a) = 2$ . Therefore

y - f (a) = f^{'} (a) (x - a), y - 1 = 2 (x - \frac{π}{4}) .

For $g (x) = \sec x$ at zero, $g (0) = 1$ and $g^{'} (0) = 0$ , so the tangent is the horizontal line $y = 1$ . A zero slope is allowed even though the secant derivative formula has no leading minus sign.

Combine with the product and quotient rules

These formulas also apply inside products and quotients. Two complete applications are:

\frac{d}{d x} [x \tan x] = \tan x + x \sec^{2} x, \cos x \neq 0 .

\frac{d}{d x} [\frac{\tan x}{x}] = \frac{x \sec^{2} x - \tan x}{x^{2}} .

The quotient requires both $x \neq 0$ and $\cos x \neq 0$ . In each calculation, identify the original structure before selecting the product rule or quotient rule.

8. Common mistakes and corrected reasoning

Mistake 1: Dropping a leading minus sign. The correct formula is $\frac{d}{d x} [\cot x] = - \csc^{2} x$ .
Mistake 2: Confusing a square with a product. The secant derivative is $\frac{d}{d x} [\sec x] = \sec x \tan x$ ; tangent has the square derivative $\sec^{2} x$ .
Mistake 3: Treating factor order as an error. The expressions $\tan x \sec x$ and $\sec x \tan x$ are equal, so either is a correct secant derivative. Both factors are required.
Mistake 4: Interchanging the tangent and cotangent formulas. Tangent gives $\sec^{2} x$ ; cotangent gives $- \csc^{2} x$ , on their respective domains.
Mistake 5: Losing a double negative. The constant multiple rule gives $\frac{d}{d x} [- \cot x] = - (- \csc^{2} x) = \csc^{2} x$ .
Mistake 6: Applying the radian formulas to a degree-valued input without rescaling. The standard formulas on this page use radians; a degree-valued variable introduces an extra scale factor.
Mistake 7: Dropping a constant coefficient. For example, $\frac{d}{d x} [5 \tan x] = 5 \sec^{2} x$ .
Mistake 8: Mixing the reciprocal-function names. Keep the pairs together: $\frac{d}{d x} [\sec x] = \sec x \tan x$ and $\frac{d}{d x} [\csc x] = - \csc x \cot x$ .

9. Strategies for learning and applying the formulas

Essential Tips

Remember formula structure: Cosine, cotangent and cosecant have explicit leading minus signs; secant and cosecant have product-form derivatives. Numerical signs still depend on the input.
Practice sign discipline: Write out negative signs explicitly
Use flashcards: These four must be instant recall for AP® exam
Double-check negatives: When you see minus signs, trace through carefully
Know the proofs: Understanding WHY helps remember the formulas
Connect to identities: $1 + \tan^{2} x = \sec^{2} x$ and $1 + \cot^{2} x = \csc^{2} x$
Radians only: Always work in radians, never degrees

Memorization Strategy

5-Step Memorization Process:

Learn the formula pattern: The standard cosine, cotangent and cosecant formulas contain leading minus signs.
Group by type: Squared (tan, cot) vs. Product (sec, csc)
Connect to sin/cos: All four derive from sin and cos using Quotient Rule
Practice sign drills: Write each derivative 10 times focusing on signs
Test yourself: Random flash cards until automatic

Radian measure and a composition preview

If $T (t) = \tan (\frac{π t}{180})$ represents a degree-valued numerical input, the chain rule gives $T^{'} (t) = \frac{π}{180} \sec^{2} (\frac{π t}{180})$ , wherever the cosine denominator is nonzero. This scale factor explains why the standard radian formula cannot be applied unchanged to a degree-valued variable.

For the later composition examples, $\frac{d}{d x} [\tan (3 x)] = 3 \sec^{2} (3 x)$ and $\frac{d}{d x} [\sec (x^{2})] = 2 x \sec (x^{2}) \tan (x^{2})$ . Keep the inner derivative and the domain of the original composition.

10. Six practice problems with solutions

Try all six original tasks before opening the solutions. Use radians and state the original domain.

1. Tangent and cosine

Differentiate $f (x) = 4 \tan x + \cos x$ .

Show the two basic derivatives

f^{'} (x) = 4 \sec^{2} x - \sin x .

The coefficient four remains, and cosine supplies the minus sign. The domain requires $\cos x \neq 0$ .

2. A negative cotangent coefficient

Differentiate $g (x) = - 3 \cot x + x^{2}$ .

Show the double negative

g^{'} (x) = - 3 (- \csc^{2} x) + 2 x = 3 \csc^{2} x + 2 x .

The cotangent factor requires $\sin x \neq 0$ .

3. Secant and cosecant

Differentiate $h (x) = \sec x + \csc x$ .

Show the two product-form derivatives

h^{'} (x) = \sec x \tan x - \csc x \cot x .

Both $\sin x \neq 0$ and $\cos x \neq 0$ are required.

4. Mixed terms

Differentiate $k (x) = 2 \tan x - 5 \sec x + 3 x$ .

Show each coefficient and sign

k^{'} (x) = 2 \sec^{2} x - 5 \sec x \tan x + 3 .

Differentiate term by term. The domain requires $\cos x \neq 0$ .

5. Cotangent, cosecant and sine

Differentiate $f (x) = \cot x + \csc x + \sin x$ .

Show both leading minus signs

f^{'} (x) = - \csc^{2} x - \csc x \cot x + \cos x .

Cotangent and cosecant each supply a leading minus sign. The domain requires $\sin x \neq 0$ .

6. Evaluate at a special angle

For $f (x) = \tan x$ , find $f^{'} (\frac{π}{4})$ .

Show the exact value and domain check

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

f^{'} (\frac{π}{4}) = \sec^{2} (\frac{π}{4}) = (\sqrt{2})^{2} = 2 .

The angle belongs to the original domain because $\cos (\frac{π}{4}) = \frac{\sqrt{2}}{2} \neq 0$ . The function value is one, while its derivative value is two.

11. Write a clear AP-style response

How to write a clear AP-style solution:

Recall the formulas fluently: Practise the four derivatives and know how quotient identities justify them.
Track signs accurately: Retain leading minus signs, coefficients and any signs from other differentiation rules.
Show the requested reasoning: Include enough differentiation and evaluation to support the result. If asked to derive a formula, show the quotient-rule and identity steps.
Simplify purposefully: Combine or factor terms when helpful or requested. Equivalent correct derivative forms can both be valid.
Watch for trick questions: Double negatives, mixed terms
Know all six trig derivatives: Often mixed with sin/cos in problems
Unit 3: Topic 3.1 introduces the chain rule for compositions. A trigonometric function with an inner expression requires that inner derivative factor.
Use proper notation: $f^{'} (x)$ , $\frac{d y}{d x}$ , etc.

Common FRQ Formats:

Find $f^{'} (x)$ for a function involving trigonometric terms.
Find the tangent slope at $x = a$ by evaluating $f^{'} (a)$ at an input in the original domain.
"Show that the derivative equals..." (prove given result)
Solve $f^{'} (x) = 0$ within the original domain. Later critical-number questions also check domain points where a derivative fails to exist.
"Find an equation of the tangent line..." (use derivative as slope)
Determine increasing intervals by analysing where $f^{'} (x) > 0$ within the original domain.
Evaluate a limit that matches a known derivative definition. Later, Topic 4.7 introduces L'Hôpital's rule for appropriate indeterminate limits in both AB and BC; it is a separate method with required conditions.

Connections to practise:

Chain Rule: $\tan (3 x)$ , $\sec (x^{2})$
Product Rule: $x \tan x$ , $e^{x} \sec x$
Quotient Rule: $\frac{\tan x}{x}$ , $\frac{x}{\cot x}$
Related rates: Problems involving trig functions changing over time
Optimization: Max/min problems with trig constraints

12. Four-formula quick reference

THE FOUR DERIVATIVES - MEMORIZE NOW!

$\frac{d}{d x} [\tan x] = \sec^{2} x$	$\frac{d}{d x} [\cot x] = - \csc^{2} x$
$\frac{d}{d x} [\sec x] = \sec x \tan x$	$\frac{d}{d x} [\csc x] = - \csc x \cot x$

Formula pattern: cotangent and cosecant have explicit leading minus signs.

Tangent and cotangent: squared forms. Secant and cosecant: product forms.

Conditions: Use radians. Tangent and secant require $\cos x \neq 0$ ; cotangent and cosecant require $\sin x \neq 0$ . Leading minus signs describe formula structure; evaluate the complete expression to determine the numerical sign.

13. Connections and next study steps

Topic 2.10 connects to:

Topic 2.7: The derivatives of $\sin x$ and $\cos x$ underpin all four quotient derivations.
Topics 2.8 and 2.9: The product and quotient rules combine these formulas in more complex expressions. All four formulas on this page can be derived using the quotient rule.
Topic 3.1: The chain rule extends the formulas to composite trigonometric functions.
Unit 5: Optimization and other analytical applications can involve trigonometric constraints.
Unit 4: Related rates with angles, periodic motion
Unit 6: Integration of trig functions (reverse process)
BC Only - Unit 10: Taylor series for trig functions use these derivatives
Real-world: Waves, oscillations, circular motion all use trig derivatives

Remember: The four formulas are $(\tan x)^{'} = \sec^{2} x$ , $(\cot x)^{'} = - \csc^{2} x$ , $(\sec x)^{'} = \sec x \tan x$ and $(\csc x)^{'} = - \csc x \cot x$ . They follow from sine and cosine identities and the quotient rule. Use radians and retain domain exclusions. A leading minus sign is part of a formula, not a guarantee that its value is negative at every input.

← Topic 2.9: quotient rule Review sine and cosine derivatives →Review the product rule →All Unit 2 study notes →

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