Lesson G1
G1 Derivative as gradient, first principles and second derivatives Quiz: AQA Maths, Unit 7
20 questions
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Lesson G1, Derivative as gradient, first principles and second derivatives: 20 multiple choice questions for the AQA Maths (7357), Unit 7: Differentiation, written with Revision Ninja.
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The 20 questions
-
Geometrically, what does the derivative f'(x) represent at a point on a curve?
- Normal gradient
- Area under curve
- Tangent gradient
- Chord length
-
Which limit defines the derivative f'(x) from first principles?
- lim (f(x+h)-f(x))/x
- lim (f(x+h)-f(x))/h
- lim (f(x+h)+f(x))/h
- lim (f(x)-f(h))/h
-
Using first principles, what is the derivative of x^2?
- 2x
- x
- 2
- x^2/2
-
What is the second derivative of y = x^3?
- 6
- 6x
- 3x
- 3x^2
-
What does the second derivative f''(x) measure geometrically?
- Rate of gradient change
- Gradient of the curve
- Area under the curve
- Rate of function change
-
What changes at a point of inflection on a curve?
- Y-intercept
- Domain
- Gradient sign
- Concavity
-
What is the derivative of x^3 with respect to x?
- 3x
- x^3/3
- 3x^2
- x^2
-
What is the gradient of y = x^2 at x = 3?
- 6
- 9
- 5
- 3
-
If f(x) = x^3, what is f'(2)?
- 6
- 8
- 4
- 12
-
Find f''(x) for f(x) = 4x^3 - 2x.
- 24x
- 12x^2 - 2
- 24
- 12x
-
What is the gradient of the curve y = x^2 - 4x at x = 1?
- -4
- -2
- 2
- -3
-
Using first principles on f(x) = x^2, what does (f(x + h) - f(x))/h simplify to before h tends to 0?
- x + h
- 2x - h
- 2x + h
- 2x
-
For f(x) = x^4 - 2x^2, what is f''(x)?
- 4x^3 - 4x
- 12x^2 - 2
- 12x^2 - 4
- 12x - 4
-
If f''(x) > 0 on an interval, what shape is the curve there?
- Convex
- Discontinuous
- Concave
- Linear
-
If f'(x) is increasing on an interval, what is the sign of f''(x) on that interval?
- Zero
- Undefined
- Negative
- Positive
-
Find the second derivative of y = 1/x^2.
- -2/x^3
- 2/x^3
- 6/x^4
- -6/x^4
-
A curve has second derivative y'' = 6x - 12. What is the x-coordinate of its point of inflection?
- 0
- 2
- -2
- 6
-
Using first principles, find f'(1) for f(x) = 3x^2 + x.
- 6
- 7
- 4
- 8
-
Why does f(x) = x^3 have a point of inflection at x = 0?
- f''(x) is positive
- f'(x) changes sign
- f(x) is zero
- f''(x) changes sign
-
Estimate the gradient of y = x^2 at x = 1 using the chord from x = 1 to x = 1.01, to 2 decimal places.
- 2.50
- 2.10
- 1.99
- 2.01
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