$$\frac{d}{dx}[u(x)v(x)] = u’(x)v(x) + u(x)v’(x)$$
Example: Differentiate $f(x) = x^2 \sin(3x)$.
$u = x^2,\ u’ = 2x;\ v = \sin(3x),\ v’ = 3\cos(3x)$.
$$f’(x) = 2x\sin(3x) + 3x^2\cos(3x)$$
$$\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \frac{u’v - uv’}{v^2}$$
Example: Differentiate $f(x) = \dfrac{e^x}{x^2+1}$.
$u = e^x,\ u’ = e^x;\ v = x^2+1,\ v’ = 2x$.
$$f’(x) = \frac{e^x(x^2+1) - 2xe^x}{(x^2+1)^2} = \frac{e^x(x^2 - 2x + 1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2}$$
$$\frac{d}{dx}[f(g(x))] = f’(g(x)) \cdot g’(x)$$
Example 1: $y = \ln(x^2+3)$.
$f(u) = \ln u,\ g(x) = x^2+3$: $y’ = \dfrac{1}{x^2+3} \cdot 2x = \dfrac{2x}{x^2+3}$.
Example 2: $y = e^{\sin x}$.
$y’ = e^{\sin x} \cdot \cos x$.
Example 3 (chain $+$ product): $y = x^2 \ln(\cos x)$.
$$y’ = 2x\ln(\cos x) + x^2 \cdot \frac{-\sin x}{\cos x} = 2x\ln(\cos x) - x^2\tan x$$
Example: $f(x) = \dfrac{x^3 - 2x}{x^2 - 1}$.
Use quotient rule, or first divide: $f(x) = x - \dfrac{x}{x^2-1}$.
Then differentiate term by term:
$$f’(x) = 1 - \frac{(x^2-1) - x(2x)}{(x^2-1)^2} = 1 - \frac{-x^2-1}{(x^2-1)^2} = 1 + \frac{x^2+1}{(x^2-1)^2}$$
These arise frequently in Specialist Mathematics:
$$\frac{d}{dx}[\arcsin(x/a)] = \frac{1}{\sqrt{a^2-x^2}}$$
$$\frac{d}{dx}[\arctan(x/a)] = \frac{a}{a^2+x^2}$$
Example: Differentiate $y = \arctan(2x)$.
$y’ = \dfrac{2}{1+(2x)^2} = \dfrac{2}{1+4x^2}$ (chain rule applied).
When $y$ is defined implicitly by $F(x,y) = 0$, differentiate both sides and solve for $dy/dx$.
Example: Find $dy/dx$ for $x^2 + y^3 = 3xy$.
$$2x + 3y^2 \frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$$
$$\frac{dy}{dx}(3y^2 - 3x) = 3y - 2x$$
$$\frac{dy}{dx} = \frac{3y - 2x}{3y^2 - 3x} = \frac{3y-2x}{3(y^2-x)}$$
For complicated products/quotients or functions of the form $f(x)^{g(x)}$:
Take $\ln$ of both sides, then differentiate.
Example: $y = x^x$. Then $\ln y = x\ln x$, so:
$$\frac{y’}{y} = \ln x + 1 \Rightarrow y’ = x^x(\ln x + 1)$$
KEY TAKEAWAY: The product, quotient, and chain rules are the three pillars of differentiation. Almost every derivative in Specialist Mathematics is found by combining these three rules.
EXAM TIP: For Paper 1 (no CAS), practise differentiating by hand until each step is automatic. Show chain rule applications clearly: write the outer derivative and the inner derivative as separate factors.
COMMON MISTAKE: In the quotient rule, writing $(uv’ - vu’)/v^2$ instead of $(u’v - uv’)/v^2$. Use the mnemonic: “lo d-hi minus hi d-lo, over lo squared.”
