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Calculus - The Concept of the Derivative (Differentiation from First Principles)

Grade 11IB

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

πŸ”‘Concepts

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The Average Rate of Change is defined as the gradient of a secant line that passes through two points on a curve, (x,f(x))(x, f(x)) and (x+h,f(x+h))(x+h, f(x+h)). Visually, this is the slope of a straight line connecting two distinct locations on a graph, representing how much the function changes over the interval hh.

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The Instantaneous Rate of Change is the gradient of the curve at one specific point. Visually, this is the slope of the tangent lineβ€”a line that just touches the curve at that point. Differentiation is the mathematical tool used to calculate this exact slope.

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Differentiation from First Principles is the formal algebraic process of finding the derivative by taking the limit of the secant gradient. As the horizontal distance hh between two points on a graph shrinks toward zero, the secant line rotates and eventually coincides with the tangent line at the point xx.

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The concept of the Limit (h→0h \to 0) is central to calculus. In a visual sense, it describes the behavior of the gradient as the 'run' of our slope calculation becomes infinitely small. We cannot simply divide by zero, so we use the limit to observe the value the gradient approaches as hh vanishes.

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Derivative Notation: The derivative of a function f(x)f(x) is commonly written as fβ€²(x)f'(x) or dydx\frac{dy}{dx}. Geometrically, if you were to plot fβ€²(x)f'(x), the yy-value of this new graph at any point xx would represent the steepness (slope) of the original function f(x)f(x) at that same xx coordinate.

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Tangents and Normals: The tangent is a line with the gradient m=fβ€²(x)m = f'(x). The normal is a line perpendicular to the tangent at the point of contact. Visually, the normal line stands at a 90∘90^{\circ} angle to the curve; its gradient is the negative reciprocal of the tangent's gradient, calculated as βˆ’1fβ€²(x)-\frac{1}{f'(x)}.

πŸ“Formulae

Gradient of a secant: m=f(x+h)βˆ’f(x)hm = \frac{f(x+h) - f(x)}{h}

The Derivative from First Principles: fβ€²(x)=lim⁑hβ†’0f(x+h)βˆ’f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Leibniz Notation: dydx=lim⁑Δxβ†’0Ξ”yΞ”x\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}

Equation of a Tangent Line: yβˆ’y1=fβ€²(x1)(xβˆ’x1)y - y_1 = f'(x_1)(x - x_1)

Gradient of a Normal Line: mnormal=βˆ’1fβ€²(x1)m_{normal} = -\frac{1}{f'(x_1)}

πŸ’‘Examples

Problem 1:

Use differentiation from first principles to find the derivative of f(x)=x2+5f(x) = x^2 + 5.

Solution:

Step 1: Substitute the function into the first principles formula: fβ€²(x)=lim⁑hβ†’0(x+h)2+5βˆ’(x2+5)hf'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 5 - (x^2 + 5)}{h} Step 2: Expand the (x+h)2(x+h)^2 term: fβ€²(x)=lim⁑hβ†’0x2+2xh+h2+5βˆ’x2βˆ’5hf'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 5 - x^2 - 5}{h} Step 3: Simplify the numerator by cancelling terms (x2x^2 and 55): fβ€²(x)=lim⁑hβ†’02xh+h2hf'(x) = \lim_{h \to 0} \frac{2xh + h^2}{h} Step 4: Factor out hh from the numerator and divide: fβ€²(x)=lim⁑hβ†’0h(2x+h)h=lim⁑hβ†’0(2x+h)f'(x) = \lim_{h \to 0} \frac{h(2x + h)}{h} = \lim_{h \to 0} (2x + h) Step 5: Apply the limit by letting h=0h = 0: fβ€²(x)=2xf'(x) = 2x

Explanation:

This approach shows that for every point on the parabola y=x2+5y = x^2 + 5, the slope of the tangent line is exactly twice the xx-coordinate.

Problem 2:

Find the gradient of the tangent to the curve f(x)=3x2f(x) = 3x^2 at the point where x=2x = 2 using first principles.

Solution:

Step 1: Set up the limit for fβ€²(x)f'(x): fβ€²(x)=lim⁑hβ†’03(x+h)2βˆ’3x2hf'(x) = \lim_{h \to 0} \frac{3(x+h)^2 - 3x^2}{h} Step 2: Expand and simplify: fβ€²(x)=lim⁑hβ†’03(x2+2xh+h2)βˆ’3x2hf'(x) = \lim_{h \to 0} \frac{3(x^2 + 2xh + h^2) - 3x^2}{h} fβ€²(x)=lim⁑hβ†’03x2+6xh+3h2βˆ’3x2hf'(x) = \lim_{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 3x^2}{h} Step 3: Divide by hh: fβ€²(x)=lim⁑hβ†’0(6x+3h)f'(x) = \lim_{h \to 0} (6x + 3h) Step 4: Evaluate the limit: fβ€²(x)=6xf'(x) = 6x Step 5: Substitute x=2x = 2 to find the specific gradient: fβ€²(2)=6(2)=12f'(2) = 6(2) = 12

Explanation:

The general derivative fβ€²(x)=6xf'(x) = 6x gives the slope at any point. By substituting x=2x=2, we find that the line touching the curve at that specific point has a very steep positive gradient of 1212.