Differentiation I

Differentiation measures how fast a quantity changes. Once the rules are automatic, the interesting work is choosing which rule a given expression needs — and often you need two of them together.

The basic rules

$$\frac{d}{dx}\left(x^n\right) = nx^{n-1}$$

Bring the power down and reduce it by one. This holds for every real $n$ — negative and fractional included.

$$\frac{d}{dx}\left(3x^4 - 5x^2 + 7\right) = 12x^3 - 10x$$

The constant 7 differentiates to 0: a constant does not change, so its rate of change is zero.

Two structural rules:

$$\frac{d}{dx}\big(f \pm g\big) = f' \pm g' \qquad \frac{d}{dx}\big(kf\big) = kf'$$

Rewrite awkward terms as powers before differentiating:

$$\frac{1}{x^2} = x^{-2} \ \longrightarrow \ -2x^{-3} \qquad \sqrt{x} = x^{1/2} \ \longrightarrow \ \tfrac12 x^{-1/2}$$

Product and quotient rules

Product rule. If $y = uv$:

$$\frac{dy}{dx} = u'v + uv'$$

Differentiate $y = x^2\sin x$.

Take $u = x^2$ and $v = \sin x$, so $u' = 2x$ and $v' = \cos x$:

$$\frac{dy}{dx} = 2x\sin x + x^2\cos x$$

Quotient rule. If $y = \dfrac{u}{v}$:

$$\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$$

Differentiate $y = \dfrac{2x+1}{x-3}$.

$u = 2x+1$, $v = x-3$, $u' = 2$, $v' = 1$:

$$\frac{dy}{dx} = \frac{2(x-3) - (2x+1)(1)}{(x-3)^2} = \frac{2x - 6 - 2x - 1}{(x-3)^2} = \frac{-7}{(x-3)^2}$$

The chain rule

For a function inside a function:

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

Differentiate $y = (3x^2+1)^5$.

Let $u = 3x^2+1$, so $y = u^5$:

$$\frac{dy}{du} = 5u^4 \qquad \frac{du}{dx} = 6x$$ $$\frac{dy}{dx} = 5(3x^2+1)^4 \cdot 6x = 30x(3x^2+1)^4$$

Informally: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.

Trigonometric derivatives

$$\frac{d}{dx}(\sin x) = \cos x \qquad \frac{d}{dx}(\cos x) = -\sin x \qquad \frac{d}{dx}(\tan x) = \sec^2 x$$

These follow from first principles using the limits $\frac{\sin\theta}{\theta} \to 1$ and $\frac{1-\cos\theta}{\theta} \to 0$ — which is why they require radians.

Combined with the chain rule:

$$\frac{d}{dx}\sin(3x) = 3\cos(3x) \qquad \frac{d}{dx}\cos(x^2) = -2x\sin(x^2)$$

Tangents and normals

$\dfrac{dy}{dx}$ evaluated at a point is the gradient of the tangent there.

Find the tangent and normal to $y = x^3 - 2x$ at $x = 2$.

The point: $y = 8 - 4 = 4$, so $(2, 4)$.

The gradient: $\dfrac{dy}{dx} = 3x^2 - 2$, which at $x = 2$ gives $10$.

Tangent:

$$y - 4 = 10(x-2) \quad \Rightarrow \quad y = 10x - 16$$

The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{10}$:

$$y - 4 = -\tfrac{1}{10}(x-2) \quad \Rightarrow \quad y = -\tfrac{1}{10}x + 4.2$$

Stationary points

A stationary point is where $\dfrac{dy}{dx} = 0$: the tangent is horizontal.

Classify it with the second derivative:

Find and classify the stationary points of $y = x^3 - 3x^2 + 4$.

$$\frac{dy}{dx} = 3x^2 - 6x = 3x(x-2) = 0 \quad \Rightarrow \quad x = 0 \text{ or } x = 2$$

$$\frac{d^2y}{dx^2} = 6x - 6$$

At $x = 0$: $\frac{d^2y}{dx^2} = -6 < 0$, a maximum. Then $y = 4$, giving $(0, 4)$.

At $x = 2$: $\frac{d^2y}{dx^2} = 6 > 0$, a minimum. Then $y = 8 - 12 + 4 = 0$, giving $(2, 0)$.

A function is increasing where $\frac{dy}{dx} > 0$ and decreasing where $\frac{dy}{dx} < 0$. For the cubic above: increasing for $x < 0$, decreasing for $0 < x < 2$, increasing again for $x > 2$ — consistent with a maximum then a minimum.

Rates of change

Connected rates use the chain rule in the form

$$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$$

The radius of a sphere increases at $0.2$ cm s$^{-1}$. Find the rate of increase of the volume when $r = 5$ cm.

$$V = \tfrac43\pi r^3 \quad \Rightarrow \quad \frac{dV}{dr} = 4\pi r^2$$

At $r = 5$: $\frac{dV}{dr} = 100\pi$.

$$\frac{dV}{dt} = 100\pi \times 0.2 = 20\pi \approx 62.8 \text{ cm}^3\text{s}^{-1}$$