Study Rates Of Change In Applied Concepts in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Calculate the derivative of f ( x ) = csc ( x ) f(x) = \csc(x) f ( x ) = csc ( x ) . Answer: f ′ ( x ) = − csc ( x ) cot ( x ) f'(x) = -\csc(x)\cot(x) f ′ ( x ) = − csc ( x ) cot ( x ) . Cosecant derivative formula.
Flashcard 2: Find the rate of change of the volume of a sphere with respect to its radius. Answer: d V d r = 4 π r 2 \frac{dV}{dr} = 4\pi r^2 d r d V = 4 π r 2 . Derivative of V = 4 3 π r 3 V = \frac{4}{3}\pi r^3 V = 3 4 π r 3 using power rule.
Flashcard 3: Evaluate the derivative of f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 at x = 1 x = 1 x = 1 . Answer: − 1 x 2 -\frac{1}{x^2} − x 2 1 , so f ′ ( 1 ) = − 1 f'(1) = -1 f ′ ( 1 ) = − 1 . Reciprocal function derivative using power rule.
Flashcard 4: Determine the rate of change for f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 at x = 5 x = 5 x = 5 . Answer: f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x , so f ′ ( 5 ) = 10 f'(5) = 10 f ′ ( 5 ) = 10 . Power rule gives 2 x 2x 2 x , substitute x = 5 x = 5 x = 5 .
Flashcard 5: What is the derivative of f ( x ) = a x f(x) = a^x f ( x ) = a x where a a a is a constant? Answer: f ′ ( x ) = a x ln ( a ) f'(x) = a^x \ln(a) f ′ ( x ) = a x ln ( a ) . Exponential derivative includes natural log factor.
Flashcard 6: What is the derivative of f ( x ) = a x f(x) = a^x f ( x ) = a x where a a a is a constant? Answer: f ′ ( x ) = a x ln ( a ) f'(x) = a^x \ln(a) f ′ ( x ) = a x ln ( a ) . Exponential derivative includes natural log factor.
Flashcard 7: Evaluate the derivative of f ( x ) = 1 x f(x) = \frac{1}{x} f ( x ) = x 1 at x = 1 x = 1 x = 1 . Answer: − 1 x 2 -\frac{1}{x^2} − x 2 1 , so f ′ ( 1 ) = − 1 f'(1) = -1 f ′ ( 1 ) = − 1 . Reciprocal function derivative using power rule.
Flashcard 8: Evaluate d d x ( x 4 ) \frac{d}{dx}(x^4) d x d ( x 4 ) at x = 1 x = 1 x = 1 . Answer: 4 x 3 4x^3 4 x 3 , so f ′ ( 1 ) = 4 f'(1) = 4 f ′ ( 1 ) = 4 . Power rule gives 4 x 3 4x^3 4 x 3 , substitute x = 1 x = 1 x = 1 .
Flashcard 9: 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 . Derivative of constant is always zero.
Flashcard 10: Determine the derivative of f ( x ) = cos ( x ) f(x) = \cos(x) f ( x ) = cos ( x ) . Answer: f ′ ( x ) = − sin ( x ) f'(x) = -\sin(x) f ′ ( x ) = − sin ( x ) . Cosine derivative is negative sine.
Flashcard 11: State the product rule for the derivative of u ( x ) v ( x ) u(x)v(x) u ( x ) v ( x ) . Answer: ( u v ) ′ = u ′ v + u v ′ (uv)' = u'v + uv' ( uv ) ′ = u ′ v + u v ′ . Product rule for multiplied functions.
Flashcard 12: Evaluate d d x ( x 3 − 2 x + 4 ) \frac{d}{dx}(x^3 - 2x + 4) d x d ( x 3 − 2 x + 4 ) at x = 0 x = 0 x = 0 . Answer: 3 x 2 − 2 3x^2 - 2 3 x 2 − 2 , so f ′ ( 0 ) = − 2 f'(0) = -2 f ′ ( 0 ) = − 2 . Differentiate then evaluate at x = 0 x = 0 x = 0 .
Flashcard 13: Evaluate d d x ( x 4 ) \frac{d}{dx}(x^4) d x d ( x 4 ) at x = 1 x = 1 x = 1 . Answer: 4 x 3 4x^3 4 x 3 , so f ′ ( 1 ) = 4 f'(1) = 4 f ′ ( 1 ) = 4 . Power rule gives 4 x 3 4x^3 4 x 3 , substitute x = 1 x = 1 x = 1 .
Flashcard 14: Determine the derivative of f ( x ) = arctan ( x ) f(x) = \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 . Arctangent derivative formula.
Flashcard 15: Identify the derivative of f ( x ) = x n f(x) = x^n f ( x ) = x n using the power rule. Answer: f ′ ( x ) = n x n − 1 f'(x) = nx^{n-1} f ′ ( x ) = n x n − 1 . Standard power rule for polynomial derivatives.
Flashcard 16: Find the rate of change of the volume of a sphere with respect to its radius. Answer: d V d r = 4 π r 2 \frac{dV}{dr} = 4\pi r^2 d r d V = 4 π r 2 . Derivative of V = 4 3 π r 3 V = \frac{4}{3}\pi r^3 V = 3 4 π r 3 using power rule.
Flashcard 17: Determine d d x ( x 2 + 3 x + 2 ) \frac{d}{dx}(x^2 + 3x + 2) d x d ( x 2 + 3 x + 2 ) . Answer: 2 x + 3 2x + 3 2 x + 3 . Sum rule and power rule applied term by term.
Flashcard 18: Find the rate of change for f ( x ) = 4 x 2 f(x) = 4x^2 f ( x ) = 4 x 2 at x = 3 x = 3 x = 3 . Answer: f ′ ( x ) = 8 x f'(x) = 8x f ′ ( x ) = 8 x , so f ′ ( 3 ) = 24 f'(3) = 24 f ′ ( 3 ) = 24 . Power rule gives 8 x 8x 8 x , substitute x = 3 x = 3 x = 3 .
Flashcard 19: What is the derivative of f ( x ) = cot ( x ) f(x) = \cot(x) f ( x ) = cot ( x ) ? Answer: f ′ ( x ) = − csc 2 ( x ) f'(x) = -\csc^2(x) f ′ ( x ) = − csc 2 ( x ) . Cotangent derivative is negative cosecant squared.
Flashcard 20: Find 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 . Arcsine derivative formula.
Flashcard 21: Identify the rate of change of y = x 3 y = x^3 y = x 3 at x = 2 x = 2 x = 2 . Answer: f ′ ( x ) = 3 x 2 f'(x) = 3x^2 f ′ ( x ) = 3 x 2 , so f ′ ( 2 ) = 12 f'(2) = 12 f ′ ( 2 ) = 12 . Apply power rule then substitute x = 2 x = 2 x = 2 .
Flashcard 22: Calculate the derivative of f ( x ) = ln ( a x ) f(x) = \ln(ax) f ( x ) = ln ( a x ) where a a a is a constant. Answer: f ′ ( x ) = 1 x f'(x) = \frac{1}{x} f ′ ( x ) = x 1 . Constant factor a a a cancels in logarithm derivative.
Flashcard 23: Evaluate d d x ( x 2 + x ) \frac{d}{dx}(x^2 + x) d x d ( x 2 + x ) at x = 2 x = 2 x = 2 . Answer: 2 x + 1 2x + 1 2 x + 1 , so f ′ ( 2 ) = 5 f'(2) = 5 f ′ ( 2 ) = 5 . Sum rule applied, then evaluate at x = 2 x = 2 x = 2 .
Flashcard 24: Calculate the derivative of f ( x ) = 2 x 3 − 3 x 2 + x f(x) = 2x^3 - 3x^2 + x f ( x ) = 2 x 3 − 3 x 2 + x . Answer: 6 x 2 − 6 x + 1 6x^2 - 6x + 1 6 x 2 − 6 x + 1 . Power rule applied to polynomial terms.
Flashcard 25: Calculate the derivative of f ( x ) = x − 1 f(x) = x^{-1} f ( x ) = x − 1 . Answer: f ′ ( x ) = − x − 2 f'(x) = -x^{-2} f ′ ( x ) = − x − 2 . Power rule applied to negative exponent.
Flashcard 26: Calculate the rate of change of f ( x ) = x 2 + 2 x f(x) = x^2 + 2x f ( x ) = x 2 + 2 x at x = 1 x = 1 x = 1 . Answer: f ′ ( x ) = 2 x + 2 f'(x) = 2x + 2 f ′ ( x ) = 2 x + 2 , so f ′ ( 1 ) = 4 f'(1) = 4 f ′ ( 1 ) = 4 . Apply power and sum rules, then evaluate.
Flashcard 27: Determine the derivative of f ( x ) = arctan ( x ) f(x) = \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 . Arctangent derivative formula.
Flashcard 28: Identify the derivative of f ( x ) = x n f(x) = x^n f ( x ) = x n using the power rule. Answer: f ′ ( x ) = n x n − 1 f'(x) = nx^{n-1} f ′ ( x ) = n x n − 1 . Standard power rule for polynomial derivatives.
Flashcard 29: Find the derivative of f ( x ) = sec ( x ) f(x) = \sec(x) f ( x ) = sec ( x ) . Answer: f ′ ( x ) = sec ( x ) tan ( x ) f'(x) = \sec(x)\tan(x) f ′ ( x ) = sec ( x ) tan ( x ) . Secant derivative formula.
Flashcard 30: Calculate the derivative of f ( x ) = csc ( x ) f(x) = \csc(x) f ( x ) = csc ( x ) . Answer: f ′ ( x ) = − csc ( x ) cot ( x ) f'(x) = -\csc(x)\cot(x) f ′ ( x ) = − csc ( x ) cot ( x ) . Cosecant derivative formula.
Flashcard 31: What is the derivative of f ( x ) = arccos ( x ) f(x) = \arccos(x) f ( x ) = arccos ( x ) ? Answer: f ′ ( x ) = − 1 1 − x 2 f'(x) = -\frac{1}{\sqrt{1-x^2}} f ′ ( x ) = − 1 − x 2 1 . Arccosine derivative is negative of arcsine.
Flashcard 32: Identify the rate of change of y = x 3 y = x^3 y = x 3 at x = 2 x = 2 x = 2 . Answer: f ′ ( x ) = 3 x 2 f'(x) = 3x^2 f ′ ( x ) = 3 x 2 , so f ′ ( 2 ) = 12 f'(2) = 12 f ′ ( 2 ) = 12 . Apply power rule then substitute x = 2 x = 2 x = 2 .
Flashcard 33: Determine the derivative of f ( x ) = cos ( x ) f(x) = \cos(x) f ( x ) = cos ( x ) . Answer: f ′ ( x ) = − sin ( x ) f'(x) = -\sin(x) f ′ ( x ) = − sin ( x ) . Cosine derivative is negative sine.
Flashcard 34: Calculate the derivative of f ( x ) = 5 x 2 − 3 x + 7 f(x) = 5x^2 - 3x + 7 f ( x ) = 5 x 2 − 3 x + 7 . Answer: 10 x − 3 10x - 3 10 x − 3 . Power rule applied to polynomial.
Flashcard 35: Determine d d x ( 7 x 2 − 4 x + 1 ) \frac{d}{dx}(7x^2 - 4x + 1) d x d ( 7 x 2 − 4 x + 1 ) . Answer: 14 x − 4 14x - 4 14 x − 4 . Apply power rule to each term.
Flashcard 36: Calculate the rate of change of f ( x ) = x 2 + 2 x f(x) = x^2 + 2x f ( x ) = x 2 + 2 x at x = 1 x = 1 x = 1 . Answer: f ′ ( x ) = 2 x + 2 f'(x) = 2x + 2 f ′ ( x ) = 2 x + 2 , so f ′ ( 1 ) = 4 f'(1) = 4 f ′ ( 1 ) = 4 . Apply power and sum rules, then evaluate.
Flashcard 37: Calculate the derivative of f ( x ) = 2 x 3 − 3 x 2 + x f(x) = 2x^3 - 3x^2 + x f ( x ) = 2 x 3 − 3 x 2 + x . Answer: 6 x 2 − 6 x + 1 6x^2 - 6x + 1 6 x 2 − 6 x + 1 . Power rule applied to polynomial terms.
Flashcard 38: Find 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 . Exponential function derivative equals itself.
Flashcard 39: Determine d d x ( 7 x 2 − 4 x + 1 ) \frac{d}{dx}(7x^2 - 4x + 1) d x d ( 7 x 2 − 4 x + 1 ) . Answer: 14 x − 4 14x - 4 14 x − 4 . Apply power rule to each term.
Flashcard 40: State the Chain Rule formula for derivatives. Answer: d y d x = d y d u ⋅ d u d x \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} d x d y = d u d y ⋅ d x d u . Composite function differentiation rule.
Flashcard 41: Find 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 . Exponential function derivative equals itself.
Flashcard 42: Find the derivative of f ( x ) = sec ( x ) f(x) = \sec(x) f ( x ) = sec ( x ) . Answer: f ′ ( x ) = sec ( x ) tan ( x ) f'(x) = \sec(x)\tan(x) f ′ ( x ) = sec ( x ) tan ( x ) . Secant derivative formula.
Flashcard 43: Find the rate of change of f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 at x = 4 x = 4 x = 4 . Answer: f ′ ( x ) = 3 x 2 f'(x) = 3x^2 f ′ ( x ) = 3 x 2 , so f ′ ( 4 ) = 48 f'(4) = 48 f ′ ( 4 ) = 48 . Power rule gives 3 x 2 3x^2 3 x 2 , substitute x = 4 x = 4 x = 4 .
Flashcard 44: What is the derivative of f ( x ) = ln ( x ) f(x) = \ln(x) f ( x ) = ln ( x ) ? Answer: f ′ ( x ) = 1 x f'(x) = \frac{1}{x} f ′ ( x ) = x 1 . Natural logarithm derivative is reciprocal.
Flashcard 45: What is the derivative of f ( x ) = ln ( x ) f(x) = \ln(x) f ( x ) = ln ( x ) ? Answer: f ′ ( x ) = 1 x f'(x) = \frac{1}{x} f ′ ( x ) = x 1 . Natural logarithm derivative is reciprocal.
Flashcard 46: Find d d x ( 3 x 3 − 5 x 2 + 4 ) \frac{d}{dx} (3x^3 - 5x^2 + 4) d x d ( 3 x 3 − 5 x 2 + 4 ) . Answer: 9 x 2 − 10 x 9x^2 - 10x 9 x 2 − 10 x . Power rule applied to each term.
Flashcard 47: Calculate the derivative of f ( x ) = sin ( x ) f(x) = \sin(x) f ( x ) = sin ( x ) . Answer: f ′ ( x ) = cos ( x ) f'(x) = \cos(x) f ′ ( x ) = cos ( x ) . Sine derivative is cosine.
Flashcard 48: What is the rate of change of y = 2 x 3 y = 2x^3 y = 2 x 3 at x = 2 x = 2 x = 2 ? Answer: f ′ ( x ) = 6 x 2 f'(x) = 6x^2 f ′ ( x ) = 6 x 2 , so f ′ ( 2 ) = 24 f'(2) = 24 f ′ ( 2 ) = 24 . Constant multiple rule with power rule.
Flashcard 49: Identify the quotient rule for u ( x ) v ( x ) \frac{u(x)}{v(x)} v ( x ) u ( x ) . Answer: ( u v ) ′ = u ′ v − u v ′ v 2 (\frac{u}{v})' = \frac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′ . Quotient rule for divided functions.
Flashcard 50: State the product rule for the derivative of u ( x ) v ( x ) u(x)v(x) u ( x ) v ( x ) . Answer: ( u v ) ′ = u ′ v + u v ′ (uv)' = u'v + uv' ( uv ) ′ = u ′ v + u v ′ . Product rule for multiplied functions.
Flashcard 51: Determine d d x ( 5 x − 1 ) \frac{d}{dx}(5x - 1) d x d ( 5 x − 1 ) at x = 1 x = 1 x = 1 . Answer: 5 5 5 , so the rate is 5 5 5 . Linear function has constant derivative.
Flashcard 52: What is the rate of change of y = 2 x 3 y = 2x^3 y = 2 x 3 at x = 2 x = 2 x = 2 ? Answer: f ′ ( x ) = 6 x 2 f'(x) = 6x^2 f ′ ( x ) = 6 x 2 , so f ′ ( 2 ) = 24 f'(2) = 24 f ′ ( 2 ) = 24 . Constant multiple rule with power rule.
Flashcard 53: Calculate the derivative of f ( x ) = ln ( a x ) f(x) = \ln(ax) f ( x ) = ln ( a x ) where a a a is a constant. Answer: f ′ ( x ) = 1 x f'(x) = \frac{1}{x} f ′ ( x ) = x 1 . Constant factor a a a cancels in logarithm derivative.
Flashcard 54: What is the derivative of f ( x ) = arccos ( x ) f(x) = \arccos(x) f ( x ) = arccos ( x ) ? Answer: f ′ ( x ) = − 1 1 − x 2 f'(x) = -\frac{1}{\sqrt{1-x^2}} f ′ ( x ) = − 1 − x 2 1 . Arccosine derivative is negative of arcsine.
Flashcard 55: Identify the quotient rule for u ( x ) v ( x ) \frac{u(x)}{v(x)} v ( x ) u ( x ) . Answer: ( u v ) ′ = u ′ v − u v ′ v 2 (\frac{u}{v})' = \frac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′ . Quotient rule for divided functions.
Flashcard 56: State the Chain Rule formula for derivatives. Answer: d y d x = d y d u ⋅ d u d x \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} d x d y = d u d y ⋅ d x d u . Composite function differentiation rule.
Flashcard 57: Determine d d x ( 5 x − 1 ) \frac{d}{dx}(5x - 1) d x d ( 5 x − 1 ) at x = 1 x = 1 x = 1 . Answer: 5 5 5 , so the rate is 5 5 5 . Linear function has constant derivative.
Flashcard 58: Evaluate d d x ( x 2 + x ) \frac{d}{dx}(x^2 + x) d x d ( x 2 + x ) at x = 2 x = 2 x = 2 . Answer: 2 x + 1 2x + 1 2 x + 1 , so f ′ ( 2 ) = 5 f'(2) = 5 f ′ ( 2 ) = 5 . Sum rule applied, then evaluate at x = 2 x = 2 x = 2 .
Flashcard 59: Find 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 . Arcsine derivative formula.
Flashcard 60: What is the derivative of f ( x ) = tan ( x ) f(x) = \tan(x) f ( x ) = tan ( x ) ? Answer: f ′ ( x ) = sec 2 ( x ) f'(x) = \sec^2(x) f ′ ( x ) = sec 2 ( x ) . Tangent derivative is secant squared.
Flashcard 61: Find the rate of change of f ( x ) = x 3 f(x) = x^3 f ( x ) = x 3 at x = 4 x = 4 x = 4 . Answer: f ′ ( x ) = 3 x 2 f'(x) = 3x^2 f ′ ( x ) = 3 x 2 , so f ′ ( 4 ) = 48 f'(4) = 48 f ′ ( 4 ) = 48 . Power rule gives 3 x 2 3x^2 3 x 2 , substitute x = 4 x = 4 x = 4 .
Flashcard 62: Determine the rate of change for f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2 at x = 5 x = 5 x = 5 . Answer: f ′ ( x ) = 2 x f'(x) = 2x f ′ ( x ) = 2 x , so f ′ ( 5 ) = 10 f'(5) = 10 f ′ ( 5 ) = 10 . Power rule gives 2 x 2x 2 x , substitute x = 5 x = 5 x = 5 .
Flashcard 63: Determine d d x ( x 2 + 3 x + 2 ) \frac{d}{dx}(x^2 + 3x + 2) d x d ( x 2 + 3 x + 2 ) . Answer: 2 x + 3 2x + 3 2 x + 3 . Sum rule and power rule applied term by term.
Flashcard 64: What is the formula for the rate of change of a function f ( x ) f(x) f ( x ) ? Answer: f ′ ( x ) f'(x) f ′ ( x ) . The derivative represents instantaneous rate of change.
Flashcard 65: Find d d x ( 3 x 3 − 5 x 2 + 4 ) \frac{d}{dx} (3x^3 - 5x^2 + 4) d x d ( 3 x 3 − 5 x 2 + 4 ) . Answer: 9 x 2 − 10 x 9x^2 - 10x 9 x 2 − 10 x . Power rule applied to each term.
Flashcard 66: What is the rate of change of the area of a circle with respect to its radius? Answer: d A d r = 2 π r \frac{dA}{dr} = 2\pi r d r d A = 2 π r . Derivative of A = π r 2 A = \pi r^2 A = π r 2 using power rule.
Flashcard 67: What is the derivative of f ( x ) = tan ( x ) f(x) = \tan(x) f ( x ) = tan ( x ) ? Answer: f ′ ( x ) = sec 2 ( x ) f'(x) = \sec^2(x) f ′ ( x ) = sec 2 ( x ) . Tangent derivative is secant squared.
Flashcard 68: What is the formula for the rate of change of a function f ( x ) f(x) f ( x ) ? Answer: f ′ ( x ) f'(x) f ′ ( x ) . The derivative represents instantaneous rate of change.
Flashcard 69: Find the rate of change for f ( x ) = 4 x 2 f(x) = 4x^2 f ( x ) = 4 x 2 at x = 3 x = 3 x = 3 . Answer: f ′ ( x ) = 8 x f'(x) = 8x f ′ ( x ) = 8 x , so f ′ ( 3 ) = 24 f'(3) = 24 f ′ ( 3 ) = 24 . Power rule gives 8 x 8x 8 x , substitute x = 3 x = 3 x = 3 .
Flashcard 70: What is the derivative of f ( x ) = cot ( x ) f(x) = \cot(x) f ( x ) = cot ( x ) ? Answer: f ′ ( x ) = − csc 2 ( x ) f'(x) = -\csc^2(x) f ′ ( x ) = − csc 2 ( x ) . Cotangent derivative is negative cosecant squared.
Flashcard 71: Calculate the derivative of f ( x ) = 5 x 2 − 3 x + 7 f(x) = 5x^2 - 3x + 7 f ( x ) = 5 x 2 − 3 x + 7 . Answer: 10 x − 3 10x - 3 10 x − 3 . Power rule applied to polynomial.
Flashcard 72: What is the rate of change of the area of a circle with respect to its radius? Answer: d A d r = 2 π r \frac{dA}{dr} = 2\pi r d r d A = 2 π r . Derivative of A = π r 2 A = \pi r^2 A = π r 2 using power rule.
Flashcard 73: Calculate the derivative of f ( x ) = x − 1 f(x) = x^{-1} f ( x ) = x − 1 . Answer: f ′ ( x ) = − x − 2 f'(x) = -x^{-2} f ′ ( x ) = − x − 2 . Power rule applied to negative exponent.
Flashcard 74: Calculate the derivative of f ( x ) = sin ( x ) f(x) = \sin(x) f ( x ) = sin ( x ) . Answer: f ′ ( x ) = cos ( x ) f'(x) = \cos(x) f ′ ( x ) = cos ( x ) . Sine derivative is cosine.
Flashcard 75: Evaluate d d x ( x 3 − 2 x + 4 ) \frac{d}{dx}(x^3 - 2x + 4) d x d ( x 3 − 2 x + 4 ) at x = 0 x = 0 x = 0 . Answer: 3 x 2 − 2 3x^2 - 2 3 x 2 − 2 , so f ′ ( 0 ) = − 2 f'(0) = -2 f ′ ( 0 ) = − 2 . Differentiate then evaluate at x = 0 x = 0 x = 0 .