| Rule | In symbols | Example |
|---|---|---|
| Constant multiple | $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(k\,\mathrm{f}(x)\right) = k\,\mathrm{f}'(x)$ | $7x^3 \to 21x^2$ |
| Sum and difference | $\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{f} \pm \mathrm{g}\right) = \mathrm{f}' \pm \mathrm{g}'$ | $x^2 + x^3 \to 2x + 3x^2$ |
| Linear term | $kx \to k$ | $-5x \to -5$ |
| Constant | $c \to 0$ | $9 \to 0$ |
Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ when $y = 4x^5 - 3x^2 + 7x - 2$.
The rule only works on terms written as $kx^n$. Roots and fractions must be rewritten using the laws of indices first.
| Written as | Rewrite as | Derivative | Tidied |
|---|---|---|---|
| $\dfrac{1}{x^2}$ | $x^{-2}$ | $-2x^{-3}$ | $-\dfrac{2}{x^3}$ |
| $\sqrt{x}$ | $x^{1/2}$ | $\tfrac12 x^{-1/2}$ | $\dfrac{1}{2\sqrt x}$ |
| $\dfrac{1}{\sqrt x}$ | $x^{-1/2}$ | $-\tfrac12 x^{-3/2}$ | $-\dfrac{1}{2x\sqrt x}$ |
| $\sqrt[3]{x^2}$ | $x^{2/3}$ | $\tfrac23 x^{-1/3}$ | $\dfrac{2}{3\sqrt[3]{x}}$ |
| $x\sqrt{x}$ | $x^{3/2}$ | $\tfrac32 x^{1/2}$ | $\tfrac32\sqrt x$ |
Differentiate $y = \dfrac{6}{x^2}$.
Differentiate $y = 4\sqrt{x}$.
Differentiate $y = \dfrac{3}{\sqrt x} + x\sqrt x$.
Differentiate $y = (2x - 1)(x + 3)$.
Differentiate $y = \dfrac{x^3 - 4x + 1}{x^2}$.
Differentiate $y = \left(\sqrt x + 2\right)^2$.
- Gradient at a known point: differentiate, then substitute the $x$-coordinate into $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ (not into $y$).
- Point with a known gradient: set $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ equal to the gradient, solve for $x$, then substitute into $y$ for the $y$-coordinate.
Find the gradient of $y = 2\sqrt x + \dfrac{8}{x}$ at the point where $x = 4$.
Find the coordinates of the points on $y = x^3 - 6x^2 + 5$ where the gradient is $-9$.
The second derivative is the rate of change of the gradient. It tells you whether the gradient is increasing or decreasing, which is how it is used to classify stationary points. In mechanics, if $s$ is displacement then $\frac{\mathrm{d}s}{\mathrm{d}t}$ is velocity and $\frac{\mathrm{d}^2s}{\mathrm{d}t^2}$ is acceleration.
(a) $y = x^4 - 2x^3 + 5x$. Find $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$. (b) $\mathrm{f}(x) = 6\sqrt x$. Find $\mathrm{f}''(4)$.
Power rule
$kx^n \to knx^{n-1}$, any rational $n$.
Constants
$c \to 0$; $kx \to k$
Sums
Differentiate term by term.
Roots
$\sqrt[m]{x^n} = x^{n/m}$
Reciprocals
$\dfrac{1}{x^n} = x^{-n}$
Products
Expand the brackets first.
Quotients
Split over a single-term denominator.
Gradient at a point
Substitute $x$ into $\dfrac{\mathrm{d}y}{\mathrm{d}x}$.
Second derivative
$\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$ or $\mathrm{f}''(x)$: differentiate again.
Differentiate $y = 5x^4 - 2x^3 + x - 9$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 20x^3 - 6x^2 + 1$
Differentiate (a) $\dfrac{3}{x}$ (b) $8\sqrt x$ (c) $\sqrt[3]{x^2}$.
▶ Show solution
(a) $3x^{-1} \to -3x^{-2} = -\dfrac{3}{x^2}$
(b) $8x^{1/2} \to 4x^{-1/2} = \dfrac{4}{\sqrt x}$
(c) $x^{2/3} \to \tfrac23 x^{-1/3}$
Find $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ when $y = (x + 3)(x - 5)$.
▶ Show solution
$y = x^2 - 2x - 15$, so $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x - 2$.
Differentiate $y = \dfrac{x^2 + 6}{\sqrt x}$.
▶ Show solution
Split: $y = \dfrac{x^2}{x^{1/2}} + \dfrac{6}{x^{1/2}} = x^{3/2} + 6x^{-1/2}$.
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = \tfrac32 x^{1/2} - 3x^{-3/2}$
Find the gradient of $y = x^3 - 2x + 4$ at the point $(2, 8)$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2$. At $x = 2$: $12 - 2 = 10$.
Given $\mathrm{f}(x) = 2x^5 - \dfrac{4}{x}$, find $\mathrm{f}'(x)$ and $\mathrm{f}''(x)$.
▶ Show solution
$\mathrm{f}(x) = 2x^5 - 4x^{-1}$
$\mathrm{f}'(x) = 10x^4 + 4x^{-2}$
$\mathrm{f}''(x) = 40x^3 - 8x^{-3}$
Find the coordinates of the points on $y = x^3 - 3x^2 - 9x + 2$ where the gradient is zero.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) = 0$
$x = 3$: $y = 27 - 27 - 27 + 2 = -25$. $x = -1$: $y = -1 - 3 + 9 + 2 = 7$.
Points $(3, -25)$ and $(-1, 7)$.
The curve $y = ax^2 + bx$ has gradient $7$ at $x = 1$ and gradient $-5$ at $x = -2$. Find $a$ and $b$.
▶ Show solution
$\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2ax + b$.
$x = 1$: $2a + b = 7$. $x = -2$: $-4a + b = -5$.
Subtract: $6a = 12$, so $a = 2$ and $b = 3$.
$\mathrm{f}(x) = 4x^{3/2} - 6x^{1/2}$ for $x > 0$. (a) Solve $\mathrm{f}'(x) = 0$. (b) Find the exact value of $\mathrm{f}''(x)$ at this value of $x$.
▶ Show solution
(a) $\mathrm{f}'(x) = 6x^{1/2} - 3x^{-1/2}$. Multiply $\mathrm{f}'(x) = 0$ by $x^{1/2}$: $6x - 3 = 0$, so $x = \tfrac12$.
(b) $\mathrm{f}''(x) = 3x^{-1/2} + \tfrac32 x^{-3/2}$.
$\left(\tfrac12\right)^{-1/2} = \sqrt2$ and $\left(\tfrac12\right)^{-3/2} = 2\sqrt2$.
$\mathrm{f}''\left(\tfrac12\right) = 3\sqrt2 + 3\sqrt2 = 6\sqrt2$.
A ball is thrown vertically upwards. Its height above the ground after $t$ seconds is modelled by $h = 20t - 5t^2$ metres.
(a) Find $\dfrac{\mathrm{d}h}{\mathrm{d}t}$ and hence the velocity after $1$ second. (b) Find when the ball is momentarily at rest, and its height then. (c) Find $\dfrac{\mathrm{d}^2h}{\mathrm{d}t^2}$ and interpret it. (d) Find the velocity with which the ball hits the ground.
▶ Show solution
(a) $\dfrac{\mathrm{d}h}{\mathrm{d}t} = 20 - 10t$. At $t = 1$: $10$ m s$^{-1}$ (upwards).
(b) $20 - 10t = 0 \Rightarrow t = 2$ s. Height $= 40 - 20 = 20$ m.
(c) $\dfrac{\mathrm{d}^2h}{\mathrm{d}t^2} = -10$: a constant acceleration of $10$ m s$^{-2}$ downwards — the model's value for gravity.
(d) $h = 0 \Rightarrow 5t(4 - t) = 0 \Rightarrow t = 4$ (not $t = 0$, which is the throw). Velocity $= 20 - 40 = -20$ m s$^{-1}$, i.e. $20$ m s$^{-1}$ downwards.