7.2 Differentiation from first principles — Practise

Work through these on your own. For every question, write the four steps in order. Show the expansion, the subtraction, the division and the limit, even when the answer feels familiar.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Differentiate $f(x) = x^2$ from first principles.
Use first principles to find the gradient of $y = x^2$ at $x = 3$ .
Differentiate $f(x) = x^2 + 2x$ from first principles.
Differentiate $f(x) = 3x^2$ from first principles.
Differentiate $f(x) = x^3$ from first principles.
Differentiate $f(x) = x^3 - x$ from first principles.
Differentiate $f(x) = x^2 + 5$ from first principles.
Differentiate $f(x) = 2x + 1$ from first principles.
Differentiate $f(x) = x^3 + x^2$ from first principles.
Show, from first principles, that the gradient of $y = x^2$ at $x = -2$ is $-4$ .

Answers and concise solutions

1. The derivative

Answer: $f(x + h) - f(x) = 2xh + h^2$ . Divide by $h$ to get $2x + h$ , then let $h \to 0$ . So $f'(x) = 2x$ .

2. The gradient at $x = 3$

Answer: Here $f(x) = x^2$ is fixed at $x = 3$ , so the chord runs from $(3, 9)$ to $(3 + h, (3 + h)^2)$ . The limit is $6$ , so the gradient at $x = 3$ is $6$ .

3. The derivative

Answer: Expanding gives $f(x + h) - f(x) = 2xh + h^2 + 2h$ . Divide by $h$ to get $2x + h + 2$ , then let $h \to 0$ . So $f'(x) = 2x + 2$ .

4. The derivative

Answer: $f(x + h) - f(x) = 6xh + 3h^2$ . Divide by $h$ to get $6x + 3h$ , then let $h \to 0$ . So $f'(x) = 6x$ , which is three times the derivative of $x^2$ .

5. The derivative

Answer: Use $(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$ . The difference is $3x^2h + 3xh^2 + h^3$ . Divide by $h$ to get $3x^2 + 3xh + h^2$ , then let $h \to 0$ . So $f'(x) = 3x^2$ .

6. The derivative

Answer: The difference is $3x^2h + 3xh^2 + h^3 - h$ . Divide by $h$ to get $3x^2 + 3xh + h^2 - 1$ , then let $h \to 0$ . So $f'(x) = 3x^2 - 1$ .

7. The derivative

Answer: The $5$ cancels in the subtraction, so the difference is $2xh + h^2$ . Dividing by $h$ and letting $h \to 0$ gives $f'(x) = 2x$ . Adding a constant moves the graph up but does not change its gradient.

8. The derivative

Answer: The difference is $2h$ . Dividing by $h$ gives $2$ , and the limit is $2$ . The derivative of a straight line is its gradient, which is $2$ everywhere.

9. The derivative

Answer: The difference is $3x^2h + 3xh^2 + h^3 + 2xh + h^2$ . Divide by $h$ to get $3x^2 + 3xh + h^2 + 2x + h$ , then let $h \to 0$ . So $f'(x) = 3x^2 + 2x$ .

10. The gradient at $x = -2$

Answer: The difference is $(4 - 4h + h^2) - 4 = -4h + h^2$ . Divide by $h$ to get $-4 + h$ , then let $h \to 0$ . The gradient is $-4$ , which is negative, so the curve is falling at $x = -2$ .