International Baccalaureate Mathematics

Syllabus Content

Informal understanding of continuity and differentiability of a function at a point.
Understanding of limits (convergence and divergence).
Definition of derivative from first principles.
Higher derivatives.

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Calculus It is said that the word calculus comes from the Latin word for the small pebble used for counting and calculations. The two major branches, differentiation and integration, are studied by pupils only towards the end of their school days but does then form a major part of their studies. A course in calculus is a prerequisite for other, more advanced courses in mathematical analysis. Isaac Newton is best known for developing calculus. He was born in England in 1642 and is considered one of the most influential scientists and mathematicians in history.

Furthermore

Official Guidance, clarification and syllabus links:

In examinations, students will not be asked to test for continuity and differentiability.

Link to: infinite geometric sequences

$$f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$

Use of this definition for polynomials only.

Familiarity with the notations $ \frac{d^ny}{dx^n}, \; f^{(n)}(x) $

Link to: proof by mathematical induction

Formula Booklet:

Derivative of $f(x)$ from first principles

$$y=f(x) \Rightarrow \frac{dy}{dx} = f'(x) = \lim_{h\to0} \left( \frac{f(x+h) - f(x)}{h} \right)$$

Differentiation from first principles is a method used to find the derivative of a function by computing the limit of the difference quotient as the change in the input variable approaches zero. In other words, it involves finding the gradient of a curve at a point by zooming in very closely and calculating the slope of the tangent line.

The key formula for differentiation from first principles is:

$$f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}$$

where $f(x)$ is the function to be differentiated.

Here is an example of how to use differentiation from first principles to find the derivative of the function $f(x) = x^2$:

$$\begin{aligned} f'(x) &= \lim_{h\to0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h\to0} \frac{(x+h)^2 - x^2}{h} \\ &= \lim_{h\to0} \frac{x^2 + 2xh + h^2 - x^2}{h} \\ &= \lim_{h\to0} \frac{2xh + h^2}{h} \\ &= \lim_{h\to0} (2x + h) \\ &= 2x \end{aligned} $$

Therefore, the derivative of $f(x) = x^2$ is $f'(x) = 2x$.

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