Differentiation
9 flashcards to master Differentiation
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What is the derivative of a function f(x), and what does it represent geometrically?
The derivative, denoted as f'(x) or dy/dx, represents the instantaneous rate of change of the function. Geometrically, it represents the gradient (slope) of the tangent line to the curve of f(x) at a specific point.
Differentiate: y = 3x^4 - 2x^2 + 5x - 7
Using the power rule (dy/dx(x^n) = nx^(n-1)), dy/dx = 12x^3 - 4x + 5. The derivative gives the gradient function of the original equation.
Explain how to find the equation of the tangent to the curve y = x^2 at the point (2, 4).
First, find the derivative: dy/dx = 2x. At x = 2, the gradient is 2(2) = 4. Then use the point-gradient form: y - 4 = 4(x - 2) giving y = 4x - 4.
What is a stationary point, and how do you find it?
A stationary point is where the gradient of the curve is zero (dy/dx = 0). To find it, differentiate the function, set the derivative equal to zero, and solve for x. Substitute x into the original equation to find the y-coordinate.
How do you determine whether a stationary point is a maximum or minimum point?
Use the second derivative test. If the second derivative (d^2y/dx^2) is positive at the stationary point, it's a minimum. If it's negative, it's a maximum. If it's zero, the test is inconclusive.
Find the coordinates of the stationary points of the curve y = x^3 - 3x
dy/dx = 3x^2 - 3. Set dy/dx = 0: 3x^2 - 3 = 0 => x = ±1. When x = 1, y = -2; when x = -1, y = 2. Stationary points are (1, -2) and (-1, 2).
A particle moves along a line such that its displacement s (in meters) from a fixed point O at time t (in seconds) is given by s = t^3 - 6t^2 + 9t. Find the times when the particle is instantaneously at rest.
The particle is at rest when its velocity is zero. Velocity, v = ds/dt = 3t^2 - 12t + 9. Setting v = 0: 3t^2 - 12t + 9 = 0 => t = 1 and t = 3 seconds.
Explain the meaning of 'rate of change' in the context of differentiation.
Rate of change describes how one variable changes in relation to another. In differentiation, dy/dx represents the rate of change of y with respect to x.
Find the maximum value of the function f(x) = -x^2 + 4x + 3.
f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. f''(x) = -2, which is negative, so x=2 is a maximum. f(2) = -4 + 8 + 3 = 7. Maximum value is 7.
Key Questions: Differentiation
What is a stationary point, and how do you find it?
A stationary point is where the gradient of the curve is zero (dy/dx = 0). To find it, differentiate the function, set the derivative equal to zero, and solve for x. Substitute x into the original equation to find the y-coordinate.
About Differentiation (2.9)
These 9 flashcards cover everything you need to know about Differentiation for your Cambridge IGCSE Mathematics (0580) exam. Each card is designed based on the official syllabus requirements.
What You'll Learn
- 1 Definitions - Key terms and their precise meanings that examiners expect
- 3 Key Concepts - Core ideas and principles from the 0580 syllabus
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After mastering Differentiation, explore these related topics:
- 2.8 Functions - 9 flashcards
- 3.1 Coordinates - 9 flashcards
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