Differentiation

What differentiation actually finds

The gradient of a straight line is the same everywhere along it. The gradient of a curve is not — it changes from point to point. Differentiation is the machinery for finding the gradient of a curve at one particular point.

That gradient is written $\dfrac{dy}{dx}$ or $f'(x)$, and it is the gradient of the tangent to the curve at that point.

From first principles

Take two points on the curve a small distance $h$ apart and find the gradient of the chord joining them. As $h$ shrinks towards zero the chord swings round until it becomes the tangent.

CXC can ask you to differentiate a simple polynomial this way, so be able to run the four steps without thinking: write $f(x+h)$, subtract $f(x)$, divide every term by $h$, then let $h \to 0$.

For $f(x) = x^2$ that gives $\dfrac{(x+h)^2 - x^2}{h} = \dfrac{2xh + h^2}{h} = 2x + h$, and as $h \to 0$ the answer is $2x$.

The power rule

Multiply by the index, then knock one off the index. It holds for every $n$ — negative and fractional included, which is the whole reason the next paragraph matters.

Rewrite before you differentiate. A term like $\dfrac{3}{x^2}$ has to become $3x^{-2}$ first, and $\sqrt{x}$ has to become $x^{1/2}$. More marks are lost in this section to differentiating before rewriting than to any other single cause.

Two more results you should not have to derive each time:

• the derivative of a constant is $0$;
• the derivative of $ax$ is $a$.

The product rule

You cannot differentiate a product by differentiating each factor and multiplying the results. Set out $u$, $u'$, $v$ and $v'$ in a small block before you touch the rule — it costs ten seconds and prevents the most common slip in the section.

Where the brackets are simple, expanding first and differentiating term by term gives the same answer and is often quicker. Use that as a check when you have time.

The quotient rule

In words: bottom times derivative of top, minus top times derivative of bottom, all over bottom squared.

The chain rule

When one function sits inside another, differentiate the outside, leave the inside exactly where it is, then multiply by the derivative of the inside.

So for $y = (3x^2 - 5)^4$: let $u = 3x^2 - 5$, giving $\dfrac{dy}{du} = 4u^3$ and $\dfrac{du}{dx} = 6x$, so $\dfrac{dy}{dx} = 24x(3x^2-5)^3$.

Tangents and normals

The gradient of the tangent at $x = a$ is $f'(a)$ — substitute into the derivative, not into the original function.

The normal is perpendicular to the tangent, so its gradient is $-\dfrac{1}{f'(a)}$. To get the equation of either line you also need the point, so substitute $x = a$ into the original $y$ as well, then use $y - y_1 = m(x - x_1)$.

Stationary points

A stationary point is a point where the tangent is flat, so $\dfrac{dy}{dx} = 0$. Solve that equation for $x$, then substitute each root back into the original $y$ to get the coordinates.

To classify each one, differentiate a second time:

Rates of change

Differentiation measures how fast one quantity changes as another changes, whatever the letters. If $s$ is displacement and $t$ is time then

$$v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$$

so velocity is the first derivative of displacement and acceleration is the second. That is the whole connection between this topic and the kinematics in Section 4 — same machinery, different letters.

"At rest" means $v = 0$, not $s = 0$. A particle at rest has stopped moving; a particle at $s = 0$ has merely returned to where it started.