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Calculus - Differentiation

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The derivative, denoted as dydx\frac{dy}{dx} or f′(x)f'(x), represents the instantaneous rate of change of a function with respect to xx.

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Geometrically, f′(a)f'(a) is the gradient (slope) of the tangent line to the curve y=f(x)y = f(x) at the point where x=ax = a.

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A stationary point occurs where the gradient is zero, i.e., f′(x)=0f'(x) = 0. These points can be local maxima, local minima, or points of horizontal inflexion.

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The normal to a curve at a given point is the line perpendicular to the tangent at that point. Its gradient mnormalm_{normal} satisfies mnormal×mtangent=−1m_{normal} \times m_{tangent} = -1.

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In IB AI, differentiation is frequently used for optimization problems, such as finding the dimensions that maximize area or minimize cost.

📐Formulae

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}

ddx(axn+bxm)=anxn−1+bmxm−1\frac{d}{dx}(ax^n + bx^m) = anx^{n-1} + bmx^{m-1}

Gradient of tangent at x=a is m=f′(a)\text{Gradient of tangent at } x=a \text{ is } m = f'(a)

Equation of tangent: y−y1=m(x−x1)\text{Equation of tangent: } y - y_1 = m(x - x_1)

Gradient of normal: mperp=−1f′(a)\text{Gradient of normal: } m_{perp} = -\frac{1}{f'(a)}

💡Examples

Problem 1:

Given the function f(x)=2x3−3x2+4f(x) = 2x^3 - 3x^2 + 4, find the coordinates of the stationary points.

Solution:

First, find the derivative: f′(x)=6x2−6xf'(x) = 6x^2 - 6x. Set the derivative to zero for stationary points: 6x2−6x=06x^2 - 6x = 0 6x(x−1)=06x(x - 1) = 0 This gives x=0x = 0 and x=1x = 1. Substitute these back into f(x)f(x) to find the yy-coordinates: For x=0x = 0, y=2(0)3−3(0)2+4=4y = 2(0)^3 - 3(0)^2 + 4 = 4. For x=1x = 1, y=2(1)3−3(1)2+4=3y = 2(1)^3 - 3(1)^2 + 4 = 3. The stationary points are (0,4)(0, 4) and (1,3)(1, 3).

Explanation:

To find stationary points, we calculate the derivative and solve for xx when f′(x)=0f'(x)=0. We then find the corresponding yy values using the original function.

Problem 2:

Find the equation of the tangent to the curve y=x2+5xy = x^2 + 5x at the point where x=2x = 2.

Solution:

  1. Find the yy-coordinate when x=2x = 2: y=(2)2+5(2)=4+10=14y = (2)^2 + 5(2) = 4 + 10 = 14. Point is (2,14)(2, 14).
  2. Find the derivative: dydx=2x+5\frac{dy}{dx} = 2x + 5
  3. Calculate the gradient (mm) at x=2x = 2: m=2(2)+5=9m = 2(2) + 5 = 9.
  4. Use the point-slope formula: y−14=9(x−2)y - 14 = 9(x - 2) y−14=9x−18y - 14 = 9x - 18 y=9x−4y = 9x - 4.

Explanation:

The gradient of the tangent is the value of the derivative at that specific xx. Once we have the gradient and the point, we use the linear equation formula.