Differentiation 2

$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) \Rightarrow 2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.

Worked example. Find $dy/dx$ for $x^2 y + y^3 = 4$.

Differentiate: $2xy + x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0$.

$\frac{dy}{dx}(x^2 + 3y^2) = -2xy$, so $dy/dx = -2xy/(x^2 + 3y^2)$.

Parametric differentiation. When $x = f(t), y = g(t)$:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.$

Worked example. $x = t^2, y = t^3$. Find $dy/dx$ at $t = 2$.

$dx/dt = 2t = 4$. $dy/dt = 3t^2 = 12$. $dy/dx = 12/4 = 3$.

For second derivative: $d^2y/dx^2 = \frac{d/dt(dy/dx)}{dx/dt}$.

• Implicit: every $y$ term gets a $dy/dx$ factor.
• Parametric: divide $t$-derivatives.
• Second derivative — differentiate again w.r.t. $t$, then divide by $dx/dt$ once more.

• Velocity: $v = ds/dt$.
• Acceleration: $a = dv/dt = d^2s/dt^2$.

Object is at rest when $v = 0$; speeding up when $v$ and $a$ have the same sign; slowing down when opposite.

Related rates. Two quantities related by an equation both change with $t$. Differentiate w.r.t. $t$ and use chain rule.

Worked example (related rates). A spherical balloon's volume increases at $20$ cm³/s. Find the rate at which the radius increases when $r = 5$.

$V = (4/3)\pi r^3 \Rightarrow dV/dt = 4\pi r^2 (dr/dt)$.

$20 = 4\pi (25)(dr/dt) \Rightarrow dr/dt = 20/(100\pi) = 1/(5\pi) \approx 0.0637$ cm/s.

Worked example (optimisation). Open rectangular box, square base. Volume $= 32$ cm³. Minimise surface area.

Base side $x$, height $h$. $x^2 h = 32 \Rightarrow h = 32/x^2$.

Surface: $S = x^2 + 4xh = x^2 + 128/x$. $S' = 2x - 128/x^2 = 0 \Rightarrow x^3 = 64 \Rightarrow x = 4$.

$S'' = 2 + 256/x^3 > 0$, so minimum. $h = 32/16 = 2$. Min surface $= 16 + 32 = 48$ cm².

• Optimisation: derivative = 0 then confirm with $f''$.
• Kinematics: derivative chain $s \to v \to a$.
• Related rates: differentiate the link equation w.r.t. $t$.