Topic 8 of 9 · Form 4
Differentiation and its applications
d/dx(xⁿ) = nxⁿ⁻¹; d/dx(eᵃˣ) = aeᵃˣ; d/dx(ln x) = 1/x; d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = sec²x. Chain, product and quotient rules.
Notes for this topic 1 file, in order
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Rules
d/dx(xⁿ) = nxⁿ⁻¹; d/dx(eᵃˣ) = aeᵃˣ; d/dx(ln x) = 1/x; d/dx(sin x) = cos x; d/dx(cos x) = −sin x; d/dx(tan x) = sec²x. Chain, product and quotient rules.
Worked example: Differentiate y = x² ln x.
dy/dx = 2x ln x + x² × 1/x = 2x ln x + x
Applications
- Gradients, tangents and normals.
- Stationary points and their nature (second derivative test).
- Connected rates of change: dy/dt = dy/dx × dx/dt.
- Small increments and approximations: δy ≈ (dy/dx)δx.
Worked example: The radius of a sphere increases at 0.5 cm/s. Find the rate of increase of volume when r = 4 cm.
V = (4/3)πr³ → dV/dr = 4πr² = 64π → dV/dt = 64π × 0.5 = 32π ≈ 101 cm³/s
Practical maximum and minimum problems
Form an expression for the quantity, differentiate, set equal to zero, and check the nature.
Worked example: A rectangle has perimeter 40 m. Find the maximum area.
A = x(20 − x) → dA/dx = 20 − 2x = 0 → x = 10 → A = 100 m² (d²A/dx² = −2 < 0, maximum)
Check yourself
- Differentiate (3x + 1)⁵.
- Differentiate sin 2x/x.
- Find the equation of the tangent to y = x² + 3x at x = 1.
Show answers
- 15(3x + 1)⁴
- (2x cos 2x − sin 2x)/x²
- Gradient 5, point (1, 4): y = 5x − 1
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