Study Introducing Calculus 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 ) = a x f(x) = a^x f ( x ) = a x , where a > 0 a > 0 a > 0 ? Answer: f ′ ( x ) = a x ln ( a ) f'(x) = a^x \text{ln}(a) f ′ ( x ) = a x ln ( a ) . For exponential with base a a a , multiply by ln ( a ) \ln(a) ln ( a ) .
Flashcard 2: What is the Quotient Rule for differentiation? Answer: If u v \frac{u}{v} v u , then v u ′ − u v ′ v 2 \frac{vu' - uv'}{v^2} v 2 v u ′ − u v ′ . For quotients: bottom times top's derivative minus top times bottom's derivative, all over bottom squared.
Flashcard 3: State the limit definition of a derivative. Answer: f ′ ( x ) = d d x f ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f'(x) = \frac{d}{dx}f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} f ′ ( x ) = d x d f ( x ) = lim h → 0 h f ( x + h ) − f ( x ) . This is the formal limit definition using the difference quotient.
Flashcard 4: Identify the derivative of 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: − 1 x − 2 -1x^{-2} − 1 x − 2 .
Flashcard 5: Differentiate f ( x ) = arccsc ( x ) f(x) = \text{arccsc}(x) f ( x ) = arccsc ( x ) . Answer: f ′ ( x ) = − 1 ∣ x ∣ sqrt ( x 2 − 1 ) f'(x) = -\frac{1}{|x|\text{sqrt}(x^2-1)} f ′ ( x ) = − ∣ x ∣ sqrt ( x 2 − 1 ) 1 . Similar to arcsec but with negative sign.
Flashcard 6: Apply the Product Rule to f ( x ) = x 2 sin ( x ) f(x) = x^2 \text{sin}(x) f ( x ) = x 2 sin ( x ) . Answer: f ′ ( x ) = 2 x sin ( x ) + x 2 cos ( x ) f'(x) = 2x\text{sin}(x) + x^2\text{cos}(x) f ′ ( x ) = 2 x sin ( x ) + x 2 cos ( x ) . Apply product rule: u = x 2 u = x^2 u = x 2 , v = sin ( x ) v = \sin(x) v = sin ( x ) .
Flashcard 7: What is the derivative of a constant function f ( x ) = c f(x) = c f ( x ) = c ? Answer: f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 . Constants have no change, so their rate of change is zero.
Flashcard 8: Apply the Chain Rule to f ( x ) = ( 3 x 2 + 2 ) 5 f(x) = (3x^2 + 2)^5 f ( x ) = ( 3 x 2 + 2 ) 5 . Answer: f ′ ( x ) = 5 ( 3 x 2 + 2 ) 4 × 6 x f'(x) = 5(3x^2 + 2)^4 \times 6x f ′ ( x ) = 5 ( 3 x 2 + 2 ) 4 × 6 x . Outer function derivative is 5 ( 3 x 2 + 2 ) 4 5(3x^2 + 2)^4 5 ( 3 x 2 + 2 ) 4 , inner is 6 x 6x 6 x .
Flashcard 9: Differentiate f ( x ) = 7 x + 4 f(x) = 7x + 4 f ( x ) = 7 x + 4 . Answer: f ′ ( x ) = 7 f'(x) = 7 f ′ ( x ) = 7 . Linear term becomes its coefficient; constant disappears.
Flashcard 10: State the limit definition of a derivative. Answer: f ′ ( x ) = d d x f ( x ) = lim h → 0 f ( x + h ) − f ( x ) h f'(x) = \frac{d}{dx}f(x) = \text{lim}_{h \to 0} \frac{f(x+h) - f(x)}{h} f ′ ( x ) = d x d f ( x ) = lim h → 0 h f ( x + h ) − f ( x ) . This is the formal limit definition using the difference quotient.
Flashcard 11: Identify 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 . Using the power rule: bring down the exponent and subtract 1.
Flashcard 12: Differentiate f ( x ) = e − x 2 f(x) = \text{e}^{-x^2} f ( x ) = e − x 2 using the Chain Rule. Answer: f ′ ( x ) = − 2 x e − x 2 f'(x) = -2x\text{e}^{-x^2} f ′ ( x ) = − 2 x e − x 2 . Chain rule: e − x 2 e^{-x^2} e − x 2 times derivative of − x 2 -x^2 − x 2 .
Flashcard 13: Apply the Product Rule to f ( x ) = x 2 sin ( x ) f(x) = x^2 \text{sin}(x) f ( x ) = x 2 sin ( x ) . Answer: f ′ ( x ) = 2 x sin ( x ) + x 2 cos ( x ) f'(x) = 2x\text{sin}(x) + x^2\text{cos}(x) f ′ ( x ) = 2 x sin ( x ) + x 2 cos ( x ) . Apply product rule: u = x 2 u = x^2 u = x 2 , v = sin ( x ) v = \sin(x) v = sin ( x ) .
Flashcard 14: Differentiate f ( x ) = x 3 − 5 x + 4 f(x) = x^3 - 5x + 4 f ( x ) = x 3 − 5 x + 4 . Answer: f ′ ( x ) = 3 x 2 − 5 f'(x) = 3x^2 - 5 f ′ ( x ) = 3 x 2 − 5 . Apply power rule term by term: 3 x 2 − 5 + 0 3x^2 - 5 + 0 3 x 2 − 5 + 0 .
Flashcard 15: Differentiate 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 . This is the standard derivative formula for inverse tangent.
Flashcard 16: What is the derivative of f ( x ) = a x f(x) = a^x f ( x ) = a x , where a > 0 a > 0 a > 0 ? Answer: f ′ ( x ) = a x ln ( a ) f'(x) = a^x \text{ln}(a) f ′ ( x ) = a x ln ( a ) . For exponential with base a a a , multiply by ln ( a ) \ln(a) ln ( a ) .
Flashcard 17: What is the Chain Rule for differentiation? Answer: If y = f ( g ( x ) ) y = f(g(x)) y = f ( g ( x )) , then d y d x = f ′ ( g ( x ) ) g ′ ( x ) \frac{dy}{dx} = f'(g(x))g'(x) d x d y = f ′ ( g ( x )) g ′ ( x ) . This rule handles composite functions: outer derivative times inner.
Flashcard 18: Identify the derivative of 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 . Similar to arctan but with negative sign.
Flashcard 19: Differentiate f ( x ) = ln ( 3 x ) f(x) = \text{ln}(3x) f ( x ) = ln ( 3 x ) using the Chain Rule. Answer: f ′ ( x ) = 3 3 x = 1 x f'(x) = \frac{3}{3x} = \frac{1}{x} f ′ ( x ) = 3 x 3 = x 1 . Chain rule: derivative of ln ( u ) \ln(u) ln ( u ) is u ′ u \frac{u'}{u} u u ′ .
Flashcard 20: Differentiate 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 ) . The derivative of sine is cosine.
Flashcard 21: Apply the Power Rule to 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 . Using power rule: 5 x 5 − 1 = 5 x 4 5x^{5-1} = 5x^4 5 x 5 − 1 = 5 x 4 .
Flashcard 22: Differentiate 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 ) . The derivative of tangent is secant squared.
Flashcard 23: 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 ) . The derivative of cotangent is negative cosecant squared.
Flashcard 24: Differentiate f ( x ) = 7 x + 4 f(x) = 7x + 4 f ( x ) = 7 x + 4 . Answer: f ′ ( x ) = 7 f'(x) = 7 f ′ ( x ) = 7 . Linear term becomes its coefficient; constant disappears.
Flashcard 25: What is the derivative of f ( x ) = arcsec ( x ) f(x) = \text{arcsec}(x) f ( x ) = arcsec ( x ) ? Answer: f ′ ( x ) = 1 ∣ x ∣ x 2 − 1 f'(x) = \frac{1}{|x| \sqrt{x^2-1}} f ′ ( x ) = ∣ x ∣ x 2 − 1 1 . The derivative includes absolute value due to domain restrictions.
Flashcard 26: Differentiate f ( x ) = arccsc ( x ) f(x) = \text{arccsc}(x) f ( x ) = arccsc ( x ) . Answer: f ′ ( x ) = − 1 ∣ x ∣ sqrt ( x 2 − 1 ) f'(x) = -\frac{1}{|x|\text{sqrt}(x^2-1)} f ′ ( x ) = − ∣ x ∣ sqrt ( x 2 − 1 ) 1 . Similar to arcsec but with negative sign.
Flashcard 27: Find the derivative of f ( x ) = 3 x 3 f(x) = 3x^3 f ( x ) = 3 x 3 . Answer: f ′ ( x ) = 9 x 2 f'(x) = 9x^2 f ′ ( x ) = 9 x 2 . Apply power rule to x 3 x^3 x 3 : 3 ⋅ 3 x 2 = 9 x 2 3 \cdot 3x^2 = 9x^2 3 ⋅ 3 x 2 = 9 x 2 .
Flashcard 28: What is the derivative of 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 . This is a fundamental derivative of logarithmic functions.
Flashcard 29: What is the derivative of 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 ) . The derivative of cosine is negative sine.
Flashcard 30: What is the derivative of 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 ) . This follows from the chain rule applied to sec ( x ) = 1 cos ( x ) \sec(x) = \frac{1}{\cos(x)} sec ( x ) = c o s ( x ) 1 .
Flashcard 31: What is the definition of a derivative? Answer: The derivative is the instantaneous rate of change of a function. This is the fundamental concept of calculus.
Flashcard 32: What is the definition of a derivative? Answer: The derivative is the instantaneous rate of change of a function. This is the fundamental concept of calculus.
Flashcard 33: What is the derivative of f ( x ) = arcsec ( x ) f(x) = \text{arcsec}(x) f ( x ) = arcsec ( x ) ? Answer: f ′ ( x ) = 1 ∣ x ∣ sqrt ( x 2 − 1 ) f'(x) = \frac{1}{|x|\text{sqrt}(x^2-1)} f ′ ( x ) = ∣ x ∣ sqrt ( x 2 − 1 ) 1 . The derivative includes absolute value due to domain restrictions.
Flashcard 34: Identify the derivative of f ( x ) = e 3 x f(x) = \text{e}^{3x} f ( x ) = e 3 x . Answer: f ′ ( x ) = 3 e 3 x f'(x) = 3\text{e}^{3x} f ′ ( x ) = 3 e 3 x . Chain rule: e 3 x e^{3x} e 3 x times derivative of 3 x 3x 3 x .
Flashcard 35: What is the derivative of f ( x ) = arccos ( x ) f(x) = \text{arccos}(x) f ( x ) = arccos ( x ) ? Answer: f ′ ( x ) = − 1 sqrt ( 1 − x 2 ) f'(x) = -\frac{1}{\text{sqrt}(1-x^2)} f ′ ( x ) = − sqrt ( 1 − x 2 ) 1 . Similar to arcsin but with negative sign.
Flashcard 36: What does the derivative tell us about a function at a point? Answer: The slope of the tangent line to the function at that point. The derivative measures the steepness of the curve at any point.
Flashcard 37: Differentiate f ( x ) = x 3 − 5 x + 4 f(x) = x^3 - 5x + 4 f ( x ) = x 3 − 5 x + 4 . Answer: f ′ ( x ) = 3 x 2 − 5 f'(x) = 3x^2 - 5 f ′ ( x ) = 3 x 2 − 5 . Apply power rule term by term: 3 x 2 − 5 + 0 3x^2 - 5 + 0 3 x 2 − 5 + 0 .
Flashcard 38: Identify the derivative of 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 ) . This follows from the chain rule applied to csc ( x ) = 1 sin ( x ) \csc(x) = \frac{1}{\sin(x)} csc ( x ) = s i n ( x ) 1 .
Flashcard 39: Identify the derivative of 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: − 1 x − 2 -1x^{-2} − 1 x − 2 .
Flashcard 40: What is the derivative of a constant function f ( x ) = c f(x) = c f ( x ) = c ? Answer: f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 . Constants have no change, so their rate of change is zero.
Flashcard 41: Apply the Power Rule to 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 . Using power rule: 5 x 5 − 1 = 5 x 4 5x^{5-1} = 5x^4 5 x 5 − 1 = 5 x 4 .
Flashcard 42: What is the derivative of f ( x ) = arccos ( x ) f(x) = \text{arccos}(x) f ( x ) = arccos ( x ) ? Answer: f ′ ( x ) = − 1 sqrt ( 1 − x 2 ) f'(x) = -\frac{1}{\text{sqrt}(1-x^2)} f ′ ( x ) = − sqrt ( 1 − x 2 ) 1 . Similar to arcsin but with negative sign.
Flashcard 43: What is the derivative of 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 ) . This follows from the chain rule applied to sec ( x ) = 1 cos ( x ) \sec(x) = \frac{1}{\cos(x)} sec ( x ) = c o s ( x ) 1 .
Flashcard 44: 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 45: What is the Chain Rule for differentiation? Answer: If y = f ( g ( x ) ) y = f(g(x)) y = f ( g ( x )) , then d y d x = f ′ ( g ( x ) ) g ′ ( x ) \frac{dy}{dx} = f'(g(x))g'(x) d x d y = f ′ ( g ( x )) g ′ ( x ) . This rule handles composite functions: outer derivative times inner.
Flashcard 46: What is the derivative of f ( x ) = log a ( x ) f(x) = \text{log}_a(x) f ( x ) = log a ( x ) ? Answer: f ′ ( x ) = 1 x ln ( a ) f'(x) = \frac{1}{x \text{ln}(a)} f ′ ( x ) = x ln ( a ) 1 . For logarithm base a a a , include the factor 1 ln ( a ) \frac{1}{\ln(a)} l n ( a ) 1 .
Flashcard 47: Apply the Chain Rule to f ( x ) = ( 3 x 2 + 2 ) 5 f(x) = (3x^2 + 2)^5 f ( x ) = ( 3 x 2 + 2 ) 5 . Answer: f ′ ( x ) = 5 ( 3 x 2 + 2 ) 4 × 6 x f'(x) = 5(3x^2 + 2)^4 \times 6x f ′ ( x ) = 5 ( 3 x 2 + 2 ) 4 × 6 x . Outer function derivative is 5 ( 3 x 2 + 2 ) 4 5(3x^2 + 2)^4 5 ( 3 x 2 + 2 ) 4 , inner is 6 x 6x 6 x .
Flashcard 48: Identify the derivative of 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 . Similar to arctan but with negative sign.
Flashcard 49: What is the Quotient Rule for differentiation? Answer: If u v \frac{u}{v} v u , then v u ′ − u v ′ v 2 \frac{vu' - uv'}{v^2} v 2 v u ′ − u v ′ .. For quotients: bottom times top's derivative minus top times bottom's derivative, all over bottom squared.
Flashcard 50: What is the derivative of 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 ) . The derivative of cosine is negative sine.
Flashcard 51: Differentiate f ( x ) = arcsin ( x ) f(x) = \text{arcsin}(x) f ( x ) = arcsin ( x ) . Answer: f ′ ( x ) = 1 sqrt ( 1 − x 2 ) f'(x) = \frac{1}{\text{sqrt}(1-x^2)} f ′ ( x ) = sqrt ( 1 − x 2 ) 1 . This is the standard derivative formula for inverse sine.
Flashcard 52: Differentiate 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 . This is the standard derivative formula for inverse tangent.
Flashcard 53: Differentiate f ( x ) = arcsin ( x ) f(x) = \text{arcsin}(x) f ( x ) = arcsin ( x ) . Answer: f ′ ( x ) = 1 sqrt ( 1 − x 2 ) f'(x) = \frac{1}{\text{sqrt}(1-x^2)} f ′ ( x ) = sqrt ( 1 − x 2 ) 1 . This is the standard derivative formula for inverse sine.
Flashcard 54: Differentiate f ( x ) = ln ( 3 x ) f(x) = \text{ln}(3x) f ( x ) = ln ( 3 x ) using the Chain Rule. Answer: f ′ ( x ) = 3 3 x = 1 x f'(x) = \frac{3}{3x} = \frac{1}{x} f ′ ( x ) = 3 x 3 = x 1 . Chain rule: derivative of ln ( u ) \ln(u) ln ( u ) is u ′ u \frac{u'}{u} u u ′ .
Flashcard 55: What does the derivative tell us about a function at a point? Answer: The slope of the tangent line to the function at that point. The derivative measures the steepness of the curve at any point.
Flashcard 56: What is the Power Rule for differentiation? Answer: If f ( x ) = x n f(x) = x^n f ( x ) = x n , then f ′ ( x ) = n x n − 1 f'(x) = nx^{n-1} f ′ ( x ) = n x n − 1 .. This is the fundamental differentiation rule for polynomial terms.
Flashcard 57: Find the derivative of f ( x ) = 3 x 3 f(x) = 3x^3 f ( x ) = 3 x 3 . Answer: f ′ ( x ) = 9 x 2 f'(x) = 9x^2 f ′ ( x ) = 9 x 2 . Apply power rule to x 3 x^3 x 3 : 3 ⋅ 3 x 2 = 9 x 2 3 \cdot 3x^2 = 9x^2 3 ⋅ 3 x 2 = 9 x 2 .
Flashcard 58: Identify the derivative of 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 ) . This follows from the chain rule applied to csc ( x ) = 1 sin ( x ) \csc(x) = \frac{1}{\sin(x)} csc ( x ) = s i n ( x ) 1 .
Flashcard 59: What is the derivative of f ( x ) = log a ( x ) f(x) = \text{log}_a(x) f ( x ) = log a ( x ) ? Answer: f ′ ( x ) = 1 x ln ( a ) f'(x) = \frac{1}{x \ln(a)} f ′ ( x ) = x l n ( a ) 1 . For logarithm base a a a , include the factor 1 ln ( a ) \frac{1}{\ln(a)} l n ( a ) 1 .
Flashcard 60: What is the Product Rule for differentiation? Answer: If u ( x ) v ( x ) u(x)v(x) u ( x ) v ( x ) , then u v ′ + v u ′ uv' + vu' u v ′ + v u ′ .. For products: first times derivative of second plus second times derivative of first.
Flashcard 61: Differentiate 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 ) . The derivative of tangent is secant squared.
Flashcard 62: Identify the derivative of f ( x ) = e 3 x f(x) = \text{e}^{3x} f ( x ) = e 3 x . Answer: f ′ ( x ) = 3 e 3 x f'(x) = 3\text{e}^{3x} f ′ ( x ) = 3 e 3 x . Chain rule: e 3 x e^{3x} e 3 x times derivative of 3 x 3x 3 x .
Flashcard 63: Differentiate f ( x ) = e − x 2 f(x) = \text{e}^{-x^2} f ( x ) = e − x 2 using the Chain Rule. Answer: f ′ ( x ) = − 2 x e − x 2 f'(x) = -2x\text{e}^{-x^2} f ′ ( x ) = − 2 x e − x 2 . Chain rule: e − x 2 e^{-x^2} e − x 2 times derivative of − x 2 -x^2 − x 2 .
Flashcard 64: What is the Power Rule for differentiation? Answer: If f ( x ) = x n f(x) = x^n f ( x ) = x n , then f ′ ( x ) = n x n − 1 f'(x) = nx^{n-1} f ′ ( x ) = n x n − 1 .. This is the fundamental differentiation rule for polynomial terms.
Flashcard 65: Differentiate 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 ) . The derivative of sine is cosine.
Flashcard 66: Identify 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 . Using the power rule: bring down the exponent and subtract 1.