Geometric Bites

Kinematics uses differentiation to connect displacement, velocity and acceleration. In this example, a stone is projected vertically upwards and its displacement is given as a function of time. Differentiation allows us to find its velocity and determine when it reaches its maximum height. The Displacement Function A stone is projected vertically upwards from the ground. After t seconds, its height above the ground, s metres, is given by: s(t) = 15t − 4.9t² 0 ≤ t ≤ 4 Question (a) – Find ds/dt Differentiate the displacement function with respect to time. Working s(t) = 15t − 4.9t² ds/dt = 15 − 4.9 × 2 × t = 15 − 9.8t Answer: ds/dt = 15 − 9.8t Question (b) – Velocity at...

Two of the most important results in differential calculus are d/dx(ln x) = 1/x d/dx(aˣ) = aˣ ln a. These formulas are closely connected. One describes the derivative of the natural logarithm, while the other gives the derivative of an exponential function with positive base. Together, they reveal the deep relationship between logarithms, exponentials, inverse functions, and differentiation. 1. Derivative of the natural logarithm Begin with ln x = y. This is equivalent to eʸ = x. Differentiate both sides with respect to y: dx/dy = eʸ. Now invert this result: dy/dx = 1/eʸ. Since eʸ = x, substitute back: dy/dx = 1/x. Therefore, d/dx(ln x) = 1/x, for x > 0. So the derivative of the natural logarithm is the reciprocal of x. 2. Derivative of the exponential function aˣ Now let y = aˣ, where a > 0 and a ≠ 1. Take logarithms to base a: logₐ y = x. Now apply the change-of-...

The Maclaurin Series — A Clean Derivation Many smooth functions can be written as an infinite polynomial. When this expansion is centred at x = 0 , we obtain the Maclaurin series . This article derives the Maclaurin formula directly from repeated differentiation, showing precisely why the coefficients involve derivatives and factorials. 1) Begin with a General Power Series Suppose a function f(x) can be expressed as f(x) = a₀ + a₁x + a₂x² + a₃x³ + … + aᵣxʳ + … The constants aᵣ are real coefficients whose values we wish to determine. 2) Evaluate at x = 0 f(0) = a₀ so a₀ = f(0). 3) Differentiate Once f′(x) = a₁ + 2a₂x + 3a₃x² + … + r·aᵣxʳ⁻¹ + … Setting x = 0 eliminates all higher powers: f′(0) = a₁. Thus, a₁ = f′(0). 4) Differentiate Again f″(x) = 2·1·a₂ + 3·2·a₃x + … + r(r−1)aᵣxʳ⁻² + … Evaluate at x = 0 : f″(0) = 2! · a₂ Hence a₂ = f″(0) / 2!. 5) The General Pattern Differentiate repeatedly. After r differentiations, a...