Reading calculus off a picture

Every derivative question so far started from a rule for the function. This lesson removes that.

Given only a graph, the gradient at a point is still a definite number: for a straight section it is rise over run, exactly; for a curve it is the tangent’s slope, which you estimate. Once you can read gradients and values off a graph, the product, quotient and chain rules work exactly as before, with numbers substituted in place of expressions.

The trap is the chain rule. To differentiate f(g(x)) at x = 1 you need g(1) first, then the gradient of f at that value, not at 1. Reading f′ at the wrong place is the most common error in this question type, which appears in the calculator-free section almost every year.

Three things transfer between a function and its derivative, and they are enough to sketch either from the other.

Where f is rising, f′ is above the axis. Where f is falling, f′ is below. Where f has a turning point, f′ crosses the axis. And where f is steepest, at a point of inflection, f′ has its own turning point.

Run it backwards and you can rebuild f from f′: zeros of f′ become stationary points, the sign pattern says whether each is a maximum or minimum, and the turning points of f′ locate the inflections. Only the vertical position of f is missing, which is why these questions usually supply one point such as f(0).

The working method, twenty practice questions, ten harder ones, and the ClassPad guide.