7.2 Differentiation from first principles — Practise with help

1. $(x + h)^2 = x^2 + 2xh + h^2$.
2. $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$.
3. Divide each term by $h$: $4h + 2$.

If question 2 was hard, revise the binomial expansion in Chapter 4 before you continue.

Write $f(x + h) = (x + h)^2 + 1$, then expand the bracket. The constant $1$ appears on both sides and cancels when you subtract.

$f(x + h) - f(x) = (x^2 + 2xh + h^2 + 1) - (x^2 + 1) = 2xh + h^2$. Divide this by $h$.

Let $h \to 0$:

The constant $1$ does not change the gradient, so the answer is the same as for $x^2$.

The factor $3$ multiplies the whole bracket. Write $f(x + h) = 3(x + h)^2$, expand the bracket, then multiply each term by $3$.

$f(x + h) = 3x^2 + 6xh + 3h^2$. Subtract $3x^2$, then divide the result by $h$.

Let $h \to 0$:

The constant multiple $3$ multiplied the derivative by $3$.

Handle the two terms separately. For the term $-x$, write $-(x + h)$ and expand the bracket with its sign.

$f(x + h) = x^2 + 2xh + h^2 - x - h$. Subtract $f(x) = x^2 - x$ to leave $2xh + h^2 - h$, then divide by $h$.

Let $h \to 0$:

Each term of the function contributes its own part of the derivative.

Use the expansion $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ for the cubic term, and expand $(x + h)$ for the linear term.

After subtracting $f(x) = x^3 + x$, the difference is $3x^2h + 3xh^2 + h^3 + h$. Divide every term by $h$.

Let $h \to 0$. The terms $3xh$ and $h^2$ vanish:

What happens to the fraction $\dfrac{f(x + h) - f(x)}{h}$ if you put $h = 0$ straight away?

With $h = 0$ the numerator and the denominator are both zero. The expression $\dfrac{0}{0}$ has no value. The student must divide by $h$ and cancel it first.

Setting $h = 0$ too early gives $\dfrac{0}{0}$, which is undefined. The correct first move is to write $f(x + h) - f(x) = 2xh + h^2$, then divide by $h$ to get $2x + h$. Only then does letting $h \to 0$ give $f'(x) = 2x$, not $0$.

For a constant function, $f(x + h)$ is the same number as $f(x)$. What is their difference?

$f(x + h) - f(x) = 5 - 5 = 0$. Divide by $h$ to get $0$, then take the limit.

The graph of $f(x) = 5$ is a horizontal line. A horizontal line has gradient zero everywhere, which matches the answer.