Study Sketching Graphs Of Functions And Derivatives in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: What is the derivative of f ( x ) = tan ( x ) f(x) = \text{tan}(x) f ( x ) = tan ( x ) ? Answer: f ′ ( x ) = sec 2 ( x ) f'(x) = \text{sec}^2(x) f ′ ( x ) = sec 2 ( x ) . Derivative of tangent is secant squared.
Flashcard 2: Find the derivative of f ( x ) = a x f(x) = a^x f ( x ) = a x . Answer: f ′ ( x ) = a x ln ( a ) f'(x) = a^x \text{ln}(a) f ′ ( x ) = a x ln ( a ) . General exponential derivative includes ln ( a ) \ln(a) ln ( a ) .
Flashcard 3: Differentiate f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 . Answer: f ′ ( x ) = − 1 x 2 f'(x) = -\frac{1}{x^2} f ′ ( x ) = − x 2 1 . Rewrite as x − 1 x^{-1} x − 1 and apply power rule.
Flashcard 4: State the Power Rule for differentiation. Answer: d d x x n = n x n − 1 \frac{d}{dx}x^n = nx^{n-1} d x d x n = n x n − 1 . Multiply by exponent, reduce exponent by 1.
Flashcard 5: Calculate f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = cos ( x ) f(x) = \text{cos}(x) f ( x ) = cos ( x ) . Answer: f ′ ( x ) = − sin ( x ) f'(x) = -\text{sin}(x) f ′ ( x ) = − sin ( x ) . Derivative of cosine is negative sine.
Flashcard 6: What is the second derivative of f ( x ) = 3 x 4 f(x) = 3x^4 f ( x ) = 3 x 4 ? Answer: f ′ ′ ( x ) = 36 x 2 f''(x) = 36x^2 f ′′ ( x ) = 36 x 2 . Apply power rule twice: f ′ ( x ) = 12 x 3 f'(x) = 12x^3 f ′ ( x ) = 12 x 3 , then again.
Flashcard 7: What is f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = sin ( x ) f(x) = \text{sin}(x) f ( x ) = sin ( x ) ? Answer: f ′ ( x ) = cos ( x ) f'(x) = \text{cos}(x) f ′ ( x ) = cos ( x ) . Derivative of sine is cosine.
Flashcard 8: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = ln ( x ) f(x) = \text{ln}(x) f ( x ) = ln ( x ) . Answer: f ′ ( x ) = 1 x f'(x) = \frac{1}{x} f ′ ( x ) = x 1 . The natural logarithm's derivative is 1 x \frac{1}{x} x 1 .
Flashcard 9: Calculate f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = arccos ( x ) f(x) = \text{arccos}(x) f ( x ) = arccos ( x ) . Answer: f ′ ( x ) = − 1 √ ( 1 − x 2 ) f'(x) = -\frac{1}{\text{√}(1-x^2)} f ′ ( x ) = − √ ( 1 − x 2 ) 1 . Negative of arcsine derivative.
Flashcard 10: Identify the critical points of f ( x ) = x 3 − 3 x f(x) = x^3 - 3x f ( x ) = x 3 − 3 x . Answer: x = 0 , x = ± √ 3 3 x = 0, x = \text{±}\frac{\text{√}3}{3} x = 0 , x = ± 3 √ 3 . Set f ′ ( x ) = 3 x 2 − 3 = 0 f'(x) = 3x^2 - 3 = 0 f ′ ( x ) = 3 x 2 − 3 = 0 and solve for x.
Flashcard 11: Calculate f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = 1 2 x 4 f(x) = \frac{1}{2}x^4 f ( x ) = 2 1 x 4 . Answer: f ′ ( x ) = 2 x 3 f'(x) = 2x^3 f ′ ( x ) = 2 x 3 . Apply power rule: 4 ⋅ 1 2 ⋅ x 3 = 2 x 3 4 \cdot \frac{1}{2} \cdot x^3 = 2x^3 4 ⋅ 2 1 ⋅ x 3 = 2 x 3 .
Flashcard 12: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = csc ( x ) f(x) = \text{csc}(x) f ( x ) = csc ( x ) . Answer: f ′ ( x ) = − csc ( x ) cot ( x ) f'(x) = -\text{csc}(x)\text{cot}(x) f ′ ( x ) = − csc ( x ) cot ( x ) . Derivative involves negative cotangent cosecant.
Flashcard 13: What is the derivative of f ( x ) = arcsin ( x ) f(x) = \arcsin(x) f ( x ) = arcsin ( x ) ? Answer: f ′ ( x ) = 1 1 − x 2 f'(x) = \frac{1}{\sqrt{1 - x^2}} f ′ ( x ) = 1 − x 2 1 . Inverse trig derivative with radical denominator.
Flashcard 14: What is the derivative of f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 ? Answer: f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x . Power rule: bring down exponent, subtract 1.
Flashcard 15: What is f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = arctan ( x ) f(x) = \text{arctan}(x) f ( x ) = arctan ( x ) ? Answer: f ′ ( x ) = 1 1 + x 2 f'(x) = \frac{1}{1+x^2} f ′ ( x ) = 1 + x 2 1 . Inverse tangent derivative with squared denominator.
Flashcard 16: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = arccot ( x ) f(x) = \text{arccot}(x) f ( x ) = arccot ( x ) . Answer: f ′ ( x ) = − 1 1 + x 2 f'(x) = -\frac{1}{1+x^2} f ′ ( x ) = − 1 + x 2 1 . Negative of arctangent derivative.
Flashcard 17: What indicates a local maximum in f ′ ( x ) f'(x) f ′ ( x ) ? Answer: Change from positive to negative. Derivative sign changes from + to - at local max.
Flashcard 18: What does f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 indicate about f ( x ) f(x) f ( x ) ? Answer: Potential extrema. Horizontal tangent lines occur at critical points.
Flashcard 19: State the Quotient Rule for differentiation. Answer: d d x [ u v ] = u ′ v − u v ′ v 2 \frac{d}{dx}[\frac{u}{v}] = \frac{u'v - uv'}{v^2} d x d [ v u ] = v 2 u ′ v − u v ′ . Low d-high minus high d-low over low squared.
Flashcard 20: What is the Product Rule for differentiation? Answer: d d x [ u v ] = u ′ v + u v ′ \frac{d}{dx}[uv] = u'v + uv' d x d [ uv ] = u ′ v + u v ′ . Sum of each function times the other's derivative.
Flashcard 21: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = csc ( x ) f(x) = \text{csc}(x) f ( x ) = csc ( x ) . Answer: f ′ ( x ) = − csc ( x ) cot ( x ) f'(x) = -\text{csc}(x)\text{cot}(x) f ′ ( x ) = − csc ( x ) cot ( x ) . Derivative involves negative cotangent cosecant.
Flashcard 22: Differentiate f ( x ) = 7 x 3 − 2 x f(x) = 7x^3 - 2x f ( x ) = 7 x 3 − 2 x . Answer: f ′ ( x ) = 21 x 2 − 2 f'(x) = 21x^2 - 2 f ′ ( x ) = 21 x 2 − 2 . Apply power rule to each term separately.
Flashcard 23: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = sec ( x ) f(x) = \text{sec}(x) f ( x ) = sec ( x ) . Answer: f ′ ( x ) = sec ( x ) tan ( x ) f'(x) = \text{sec}(x)\text{tan}(x) f ′ ( x ) = sec ( x ) tan ( x ) . Product of secant and tangent functions.
Flashcard 24: What is the Product Rule for differentiation? Answer: d d x [ u v ] = u ′ v + u v ′ \frac{d}{dx}[uv] = u'v + uv' d x d [ uv ] = u ′ v + u v ′ . Sum of each function times the other's derivative.
Flashcard 25: Identify the intervals where f ( x ) = x 2 − 4 f(x) = x^2 - 4 f ( x ) = x 2 − 4 is increasing. Answer: x > 0 x > 0 x > 0 . f ′ ( x ) = 2 x > 0 f'(x) = 2x > 0 f ′ ( x ) = 2 x > 0 when x > 0 x > 0 x > 0 .
Flashcard 26: State the Chain Rule for differentiation. Answer: d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) g ′ ( x ) . Differentiate outer function times inner derivative.
Flashcard 27: State the Chain Rule for differentiation. Answer: d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x))g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) g ′ ( x ) . Differentiate outer function times inner derivative.
Flashcard 28: What is the derivative of f ( x ) = cot ( x ) f(x) = \text{cot}(x) f ( x ) = cot ( x ) ? Answer: f ′ ( x ) = − csc 2 ( x ) f'(x) = -\text{csc}^2(x) f ′ ( x ) = − csc 2 ( x ) . Derivative of cotangent is negative cosecant squared.
Flashcard 29: Differentiate f ( x ) = 7 x 3 − 2 x f(x) = 7x^3 - 2x f ( x ) = 7 x 3 − 2 x . Answer: f ′ ( x ) = 21 x 2 − 2 f'(x) = 21x^2 - 2 f ′ ( x ) = 21 x 2 − 2 . Apply power rule to each term separately.
Flashcard 30: What is f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = x 5 f(x) = x^5 f ( x ) = x 5 ? Answer: f ′ ( x ) = 5 x 4 f'(x) = 5x^4 f ′ ( x ) = 5 x 4 . Power rule: bring down 5, subtract 1 from exponent.
Flashcard 31: What is the inverse function derivative formula? Answer: [ f − 1 ] ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) [f^{-1}]'(x) = \frac{1}{f'(f^{-1}(x))} [ f − 1 ] ′ ( x ) = f ′ ( f − 1 ( x )) 1 . Reciprocal of derivative at corresponding point.
Flashcard 32: Find the critical points of f ( x ) = x 2 − 6 x + 8 f(x) = x^2 - 6x + 8 f ( x ) = x 2 − 6 x + 8 . Answer: x = 3 x = 3 x = 3 . Set f ′ ( x ) = 2 x − 6 = 0 f'(x) = 2x - 6 = 0 f ′ ( x ) = 2 x − 6 = 0 and solve.
Flashcard 33: What indicates a local maximum in f ′ ( x ) f'(x) f ′ ( x ) ? Answer: Change from positive to negative. Derivative sign changes from + to - at local max.
Flashcard 34: Which test identifies concavity? Answer: Second Derivative Test. Examines sign of f ′ ′ ( x ) f''(x) f ′′ ( x ) to determine concavity.
Flashcard 35: What is f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = arctan ( x ) f(x) = \text{arctan}(x) f ( x ) = arctan ( x ) ? Answer: f ′ ( x ) = 1 1 + x 2 f'(x) = \frac{1}{1+x^2} f ′ ( x ) = 1 + x 2 1 . Inverse tangent derivative with squared denominator.
Flashcard 36: Differentiate f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 . Answer: f ′ ( x ) = − 1 x 2 f'(x) = -\frac{1}{x^2} f ′ ( x ) = − x 2 1 . Rewrite as x − 1 x^{-1} x − 1 and apply power rule.
Flashcard 37: What is f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = sin ( x ) f(x) = \text{sin}(x) f ( x ) = sin ( x ) ? Answer: f ′ ( x ) = cos ( x ) f'(x) = \text{cos}(x) f ′ ( x ) = cos ( x ) . Derivative of sine is cosine.
Flashcard 38: State the Quotient Rule for differentiation. Answer: d d x [ u v ] = u ′ v − u v ′ v 2 \frac{d}{dx}[\frac{u}{v}] = \frac{u'v - uv'}{v^2} d x d [ v u ] = v 2 u ′ v − u v ′ . Low d-high minus high d-low over low squared.
Flashcard 39: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = arccot ( x ) f(x) = \text{arccot}(x) f ( x ) = arccot ( x ) . Answer: f ′ ( x ) = − 1 1 + x 2 f'(x) = -\frac{1}{1+x^2} f ′ ( x ) = − 1 + x 2 1 . Negative of arctangent derivative.
Flashcard 40: Which test identifies concavity? Answer: Second Derivative Test. Examines sign of f ′ ′ ( x ) f''(x) f ′′ ( x ) to determine concavity.
Flashcard 41: Identify the concavity of f ( x ) = x 4 f(x) = x^4 f ( x ) = x 4 . Answer: Concave up for all x x x . f ′ ′ ( x ) = 12 x 2 ≥ 0 f''(x) = 12x^2 \geq 0 f ′′ ( x ) = 12 x 2 ≥ 0 for all real x.
Flashcard 42: Identify the intervals where f ( x ) = x 2 − 4 f(x) = x^2 - 4 f ( x ) = x 2 − 4 is increasing. Answer: x > 0 x > 0 x > 0 . f ′ ( x ) = 2 x > 0 f'(x) = 2x > 0 f ′ ( x ) = 2 x > 0 when x > 0 x > 0 x > 0 .
Flashcard 43: Which test uses f ′ ′ ( x ) f''(x) f ′′ ( x ) to find extrema? Answer: Second Derivative Test. Uses second derivative to classify critical points.
Flashcard 44: What is the derivative of f ( x ) = cot ( x ) f(x) = \text{cot}(x) f ( x ) = cot ( x ) ? Answer: f ′ ( x ) = − csc 2 ( x ) f'(x) = -\text{csc}^2(x) f ′ ( x ) = − csc 2 ( x ) . Derivative of cotangent is negative cosecant squared.
Flashcard 45: What is the second derivative of f ( x ) = 3 x 4 f(x) = 3x^4 f ( x ) = 3 x 4 ? Answer: f ′ ′ ( x ) = 36 x 2 f''(x) = 36x^2 f ′′ ( x ) = 36 x 2 . Apply power rule twice: f ′ ( x ) = 12 x 3 f'(x) = 12x^3 f ′ ( x ) = 12 x 3 , then again.
Flashcard 46: What is the derivative of a constant c c c ? Answer: 0 0 0 . Constants have zero rate of change.
Flashcard 47: Calculate f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = cos ( x ) f(x) = \text{cos}(x) f ( x ) = cos ( x ) . Answer: f ′ ( x ) = − sin ( x ) f'(x) = -\text{sin}(x) f ′ ( x ) = − sin ( x ) . Derivative of cosine is negative sine.
Flashcard 48: Find the derivative of f ( x ) = a x f(x) = a^x f ( x ) = a x . Answer: f ′ ( x ) = a x ln ( a ) f'(x) = a^x \text{ln}(a) f ′ ( x ) = a x ln ( a ) . General exponential derivative includes ln ( a ) \ln(a) ln ( a ) .
Flashcard 49: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = ln ( x ) f(x) = \text{ln}(x) f ( x ) = ln ( x ) . Answer: f ′ ( x ) = 1 x f'(x) = \frac{1}{x} f ′ ( x ) = x 1 . The natural logarithm's derivative is 1 x \frac{1}{x} x 1 .
Flashcard 50: What is the derivative of f ( x ) = arcsin ( x ) f(x) = \text{arcsin}(x) f ( x ) = arcsin ( x ) ? Answer: f ′ ( x ) = 1 √ ( 1 − x 2 ) f'(x) = \frac{1}{\text{√}(1-x^2)} f ′ ( x ) = √ ( 1 − x 2 ) 1 . Inverse trig derivative with radical denominator.
Flashcard 51: Find the critical points of f ( x ) = x 2 − 6 x + 8 f(x) = x^2 - 6x + 8 f ( x ) = x 2 − 6 x + 8 . Answer: x = 3 x = 3 x = 3 . Set f ′ ( x ) = 2 x − 6 = 0 f'(x) = 2x - 6 = 0 f ′ ( x ) = 2 x − 6 = 0 and solve.
Flashcard 52: What does f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 indicate about f ( x ) f(x) f ( x ) ? Answer: Potential extrema. Horizontal tangent lines occur at critical points.
Flashcard 53: State the Power Rule for differentiation. Answer: d d x x n = n x n − 1 \frac{d}{dx}x^n = nx^{n-1} d x d x n = n x n − 1 . Multiply by exponent, reduce exponent by 1.
Flashcard 54: Identify the concavity of f ( x ) = x 4 f(x) = x^4 f ( x ) = x 4 . Answer: Concave up for all x x x . f ′ ′ ( x ) = 12 x 2 ≥ 0 f''(x) = 12x^2 \geq 0 f ′′ ( x ) = 12 x 2 ≥ 0 for all real x.
Flashcard 55: What is the inverse function derivative formula? Answer: [ f − 1 ] ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) [f^{-1}]'(x) = \frac{1}{f'(f^{-1}(x))} [ f − 1 ] ′ ( x ) = f ′ ( f − 1 ( x )) 1 . Reciprocal of derivative at corresponding point.
Flashcard 56: What is the derivative of f ( x ) = e x f(x) = e^x f ( x ) = e x ? Answer: f ′ ( x ) = e x f'(x) = e^x f ′ ( x ) = e x . The exponential function is its own derivative.
Flashcard 57: What is the derivative of f ( x ) = e x f(x) = e^x f ( x ) = e x ? Answer: f ′ ( x ) = e x f'(x) = e^x f ′ ( x ) = e x . The exponential function is its own derivative.
Flashcard 58: What is the derivative of a constant c c c ? Answer: 0 0 0 . Constants have zero rate of change.
Flashcard 59: Find f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = sec ( x ) f(x) = \text{sec}(x) f ( x ) = sec ( x ) . Answer: f ′ ( x ) = sec ( x ) tan ( x ) f'(x) = \text{sec}(x)\text{tan}(x) f ′ ( x ) = sec ( x ) tan ( x ) . Product of secant and tangent functions.
Flashcard 60: Calculate f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = arccos ( x ) f(x) = \arccos(x) f ( x ) = arccos ( x ) . Answer: f'(x) = -\frac{1}{\sqrt{1-x^2)} . Negative of arcsine derivative.
Flashcard 61: Identify the critical points of f ( x ) = x 3 − 3 x f(x) = x^3 - 3x f ( x ) = x 3 − 3 x . Answer: x = 0 , x = ± √ 3 3 x = 0, x = \text{±}\frac{\text{√}3}{3} x = 0 , x = ± 3 √ 3 . Set f ′ ( x ) = 3 x 2 − 3 = 0 f'(x) = 3x^2 - 3 = 0 f ′ ( x ) = 3 x 2 − 3 = 0 and solve for x.
Flashcard 62: Calculate f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = 1 2 x 4 f(x) = \frac{1}{2}x^4 f ( x ) = 2 1 x 4 . Answer: f ′ ( x ) = 2 x 3 f'(x) = 2x^3 f ′ ( x ) = 2 x 3 . Apply power rule: 4 ⋅ 1 2 ⋅ x 3 = 2 x 3 4 \cdot \frac{1}{2} \cdot x^3 = 2x^3 4 ⋅ 2 1 ⋅ x 3 = 2 x 3 .
Flashcard 63: Which test uses f ′ ′ ( x ) f''(x) f ′′ ( x ) to find extrema? Answer: Second Derivative Test. Uses second derivative to classify critical points.
Flashcard 64: What is the derivative of f ( x ) = tan ( x ) f(x) = \text{tan}(x) f ( x ) = tan ( x ) ? Answer: f ′ ( x ) = sec 2 ( x ) f'(x) = \text{sec}^2(x) f ′ ( x ) = sec 2 ( x ) . Derivative of tangent is secant squared.
Flashcard 65: What is the derivative of f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 ? Answer: f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x . Power rule: bring down exponent, subtract 1.
Flashcard 66: What is f ′ ( x ) f'(x) f ′ ( x ) for f ( x ) = x 5 f(x) = x^5 f ( x ) = x 5 ? Answer: f ′ ( x ) = 5 x 4 f'(x) = 5x^4 f ′ ( x ) = 5 x 4 . Power rule: bring down 5, subtract 1 from exponent.