Differentiation

📚The mathematics

The basic rule

Multiply by the power, then reduce the power by one. It works for negative and fractional powers too, which is why the first move on any awkward expression is to rewrite it as a power:

Most “I cannot differentiate this” moments are really “I have not rewritten this yet” moments. Note the last one: divide through first when the denominator is a single term, since there is no quotient rule at P1.

The notation \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \) and \( \mathrm{f}'(x) \) mean the same thing. Both represent the gradient of the tangent, and equivalently the rate of change of \(y\) with respect to \(x\).

The chain rule

Differentiate the outside, leave the inside alone, then multiply by the derivative of the inside. The factor people forget is that last one: the derivative of \( (3x+1)^{4} \) is \( 12(3x+1)^{3} \), not \( 4(3x+1)^{3} \).

Tangents and normals

At the point where \( x = a \):

• the \(y\)-coordinate comes from \( \mathrm{f}(a) \) — the function, not the derivative;
• the gradient of the tangent is \( \mathrm{f}'(a) \);
• the gradient of the normal is \( -\dfrac{1}{\mathrm{f}'(a)} \), since the normal is perpendicular.

Then use \( y - y_1 = m(x - x_1) \). Two errors account for nearly all lost marks here: substituting into \( \mathrm{f}' \) to get the \(y\)-coordinate, and using the tangent gradient for the normal.

Increasing, decreasing and stationary

“Show that \( \mathrm{f} \) is increasing for all \(x\)” usually means showing \( \mathrm{f}'(x) \) is a sum of squares or otherwise clearly positive — completing the square on the derivative is the standard technique.

Classifying stationary points

Solve \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \) to locate them, then classify by one of two methods:

• Second derivative. \( \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} > 0 \) means a minimum (the curve is concave up, holding water); \( \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} < 0 \) means a maximum. This is quicker when the second derivative is easy.
• Sign test. Evaluate \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \) just either side of the point and record the signs. Necessary when the second derivative is zero, which is inconclusive.

The second derivative being zero does not prove a point of inflexion — \( y = x^{4} \) has \( \dfrac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 0 \) at the origin and a clear minimum there. Fall back on the sign test.

Practical maxima and minima

These questions follow a fixed procedure, and the marks are spread across all of it:

1. Write an expression for the quantity to be optimised.
2. Use the constraint to eliminate the second variable, so you have a function of one variable. This is usually the hardest step and it is where the “show that” part often sits.
3. Differentiate and set to zero.
4. Solve, rejecting any root that makes no physical sense (a negative length, for instance).
5. Classify — show it is the maximum or minimum you claimed.
6. Answer the question actually asked, which is often the optimal value rather than the \(x\) that produces it.

Connected rates of change

When two quantities both vary with time, the chain rule links their rates:

The method: write down what you are given and what you want, both as derivatives, then find the missing link by differentiating a formula connecting the variables. Choose the chain so the intermediate quantities cancel, exactly as fractions would.