Differentiation

1.7 Differentiation

The gradient of a curve

The gradient of a curve at a point is the gradient of its tangent there. It is the limit of the gradients of chords from that point, as the other end of the chord moves closer and closer.

The power rule

Rewrite roots and fractions as powers first:

$\sqrt{x} = x^{\frac{1}{2}} \qquad \frac{3}{x^2} = 3x^{-2} \qquad \frac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$

$y = 4x^3 - 3x^{-2} + 2x^{\frac{1}{2}}$, so

$\frac{\mathrm{d}y}{\mathrm{d}x} = 12x^2 + 6x^{-3} + x^{-\frac{1}{2}}$

The chain rule

For a function of a function, differentiate the outside, keep the inside, then multiply by the derivative of the inside.

$\frac{\mathrm{d}}{\mathrm{d}x}(3x + 1)^4 = 4(3x + 1)^3 \times 3 = 12(3x + 1)^3$

$\frac{\mathrm{d}}{\mathrm{d}x}(2x + 1)^{\frac{1}{2}} = \tfrac{1}{2}(2x + 1)^{-\frac{1}{2}} \times 2 = (2x + 1)^{-\frac{1}{2}}$

Tangents and normals

The tangent has the gradient of the curve at that point. The normal is perpendicular to it.

At $x = 2$, $y = 4$ and $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 2 = 10$.

Tangent: $y - 4 = 10(x - 2)$, so $y = 10x - 16$.

Normal: the gradient is $-\tfrac{1}{10}$, so $y - 4 = -\tfrac{1}{10}(x - 2)$, which is $x + 10y = 42$.

Increasing and decreasing functions

$\mathrm{f}'(x) = 3x^2 + 6x + 6 = 3\left[(x + 1)^2 + 1\right]$. The bracket is always at least 1, so $\mathrm{f}'(x) > 0$ for all $x$.

Rates of change

When quantities change with time, connect them with the chain rule.

$A = \pi r^2$, so $\dfrac{\mathrm{d}A}{\mathrm{d}r} = 2\pi r = 10\pi$ when $r = 5$.

$\frac{\mathrm{d}A}{\mathrm{d}t} = 10\pi \times 0.2 = 2\pi \text{ cm}^2\text{ s}^{-1}$

Stationary points

A stationary point is where $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$: the tangent is horizontal.

Use the second derivative to decide which kind it is:

$\frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 - 6x - 9 = 3(x - 3)(x + 1) = 0 \quad\Rightarrow\quad x = 3 \text{ or } x = -1$

The points are $(3, -22)$ and $(-1, 10)$. The second derivative is $6x - 6$:

• At $x = 3$ it is $12 > 0$, so $(3, -22)$ is a minimum.
• At $x = -1$ it is $-12 < 0$, so $(-1, 10)$ is a maximum.

Maximum and minimum problems

Write the quantity in terms of one variable, differentiate, set to zero, and confirm the nature.

If one side is $x$, the other is $20 - x$, so $A = x(20 - x) = 20x - x^2$.

$\dfrac{\mathrm{d}A}{\mathrm{d}x} = 20 - 2x = 0$ gives $x = 10$. $\dfrac{\mathrm{d}^2A}{\mathrm{d}x^2} = -2 < 0$, so this is a maximum, and the greatest area is 100 cm².