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Flashcards · 7357 · Topic 8
G: Differentiation flashcards
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Topic 8: G: Differentiation
- The two meanings of a derivative
- Geometric: the gradient of the tangent line. Physical: the instantaneous rate of change.
- Points of inflection
- Where f″(x) = 0 and the concavity changes (f″ changes sign).
- Differentiation from first principles
- f′(x) = lim (h → 0) [f(x + h) − f(x)]/h. Required for small positive integer powers of x and for sin x and cos x.
- Standard derivatives of xⁿ, e^(kx), aᵏˣ, ln x, sin kx, cos kx and tan kx
- nxⁿ⁻¹ k e^(kx) k aᵏˣ ln a 1/x k cos kx −k sin kx k sec² kx
- Product rule and quotient rule
- y = uv: dy/dx = u dv/dx + v du/dx. y = u/v: dy/dx = (v du/dx − u dv/dx)/v².
- Chain rule, with an example
- dy/dx = (dy/du) × (du/dx). d/dx (3x² + 1)⁵ = 5(3x² + 1)⁴ × 6x = 30x(3x² + 1)⁴.
- Convex and concave
- Convex: f″(x) > 0 (the gradient is increasing). Concave: f″(x) < 0 (the gradient is decreasing).
- How do you find and classify stationary points?
- Solve f′(x) = 0. Then f″(x) > 0: minimum; f″(x) < 0: maximum. If f″(x) = 0, check the sign of f′(x) either side.
- Implicit and parametric differentiation (first derivative only)
- Implicit: differentiate both sides with respect to x; each y term gains a factor dy/dx (d/dx(y²) = 2y dy/dx). Parametric: dy/dx = (dy/dt) ÷ (dx/dt).
- Sketching a gradient function from a curve
- Where the curve increases, the gradient function is positive; where it decreases, negative; at each turning point the gradient function crosses zero.