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Calculus - Limit / derivative (introduction)

Grade 11IB_AA

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

🔑Concepts

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The limit lim⁡x→af(x)=L\lim_{x \to a} f(x) = L means that as xx approaches aa from both sides, the value of f(x)f(x) approaches LL.

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A function f(x)f(x) is continuous at x=ax = a if lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a).

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The derivative f′(x)f'(x) represents the instantaneous rate of change of a function or the gradient of the tangent to the curve at a specific point.

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The derivative from first principles is the formal definition of the derivative using limits: f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.

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The notation for the derivative includes f′(x)f'(x), dydx\frac{dy}{dx}, or ddx[f(x)]\frac{d}{dx}[f(x)].

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The Power Rule is a fundamental rule for finding derivatives of functions in the form f(x)=xnf(x) = x^n.

📐Formulae

lim⁡h→0f(x+h)−f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

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

ddx(cf(x))=c⋅f′(x)\frac{d}{dx}(cf(x)) = c \cdot f'(x)

ddx(f(x)±g(x))=f′(x)±g′(x)\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)

💡Examples

Problem 1:

Evaluate the limit: lim⁡x→4x2−16x−4\lim_{x \to 4} \frac{x^2 - 16}{x - 4}.

Solution:

lim⁡x→4(x−4)(x+4)x−4=lim⁡x→4(x+4)=4+4=8\lim_{x \to 4} \frac{(x-4)(x+4)}{x-4} = \lim_{x \to 4} (x+4) = 4 + 4 = 8.

Explanation:

Since direct substitution results in an indeterminate form 00\frac{0}{0}, we factor the numerator using the difference of squares and cancel the common factor (x−4)(x-4) before substituting x=4x = 4.

Problem 2:

Find the derivative of f(x)=x2+3xf(x) = x^2 + 3x using first principles.

Solution:

f′(x)=lim⁡h→0(x+h)2+3(x+h)−(x2+3x)hf'(x) = \lim_{h \to 0} \frac{(x+h)^2 + 3(x+h) - (x^2 + 3x)}{h} f′(x)=lim⁡h→0x2+2xh+h2+3x+3h−x2−3xhf'(x) = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 3x + 3h - x^2 - 3x}{h} f′(x)=lim⁡h→02xh+h2+3hhf'(x) = \lim_{h \to 0} \frac{2xh + h^2 + 3h}{h} f′(x)=lim⁡h→0(2x+h+3)=2x+3f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3.

Explanation:

Apply the definition of the derivative. Expand the terms, simplify the numerator by canceling like terms, divide by hh, and then evaluate the limit as hh approaches 00.

Problem 3:

Find the gradient of the curve y=2x3−5x+1y = 2x^3 - 5x + 1 at the point where x=2x = 2.

Solution:

Step 1: Find the derivative dydx=6x2−5\frac{dy}{dx} = 6x^2 - 5. \nStep 2: Substitute x=2x = 2: 6(2)2−5=6(4)−5=24−5=196(2)^2 - 5 = 6(4) - 5 = 24 - 5 = 19.

Explanation:

First, use the power rule to differentiate the function. The gradient of the curve at a specific point is the value of the derivative at that point.