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International Baccalaureate Mathematics

Calculus

Syllabus Content

Implicit differentiation.
Related rates of change.
Optimisation problems.

Here is an exam-style questions on this statement:

Here are some Advanced Starters on this statement:

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Furthermore

Official Guidance, clarification and syllabus links:

Appropriate use of the chain rule or implicit differentiation, including cases where the optimum solution is at the end point.


Implicit differentiation is a technique used to find the derivative of a function that is not written in the form of \(y = f(x)\), but rather as an equation involving \(x\) and \(y\). To use implicit differentiation, we differentiate both sides of the equation with respect to \(x\) and then solve for the derivative of \(y\) in terms of \(x\) and \(y\).

The key formula for implicit differentiation is the chain rule:

$$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx}$$

where \(t\) is a parameter that defines the curve.

Here is an example of how to use implicit differentiation to find the derivative of the equation \(x^2 + y^2 = 25\):

$$\begin{aligned} x^2 + y^2 &= 25 \\ \frac{d}{dx}(x^2 + y^2) &= \frac{d}{dx}(25) \\ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) &= 0 \\ 2x + 2y\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -\frac{x}{y} \end{aligned} $$

Therefore, the derivative of the equation \(x^2 + y^2 = 25\) with respect to \(x\) is given by \(y' = -x/y\).


Related rates of change. Here is a suggested step-by-step process for solving many problems:

  1. Label a Diagram of the situation and notice which values are variables and which are constants;
  2. Write an Equation relating the variables;
  3. Differentiate the equation with respect to time (t);
  4. Use the values at a Particular time into the differentiated equation;
  5. Make the required rate the Subject.

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