Theorem: Derivative of Parametric Functions

Theorem. If $x=f(t)$ and $y=g(t)$ are differentiable functions of $t$, then $y$ is a differentiable function of $x$, and $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt},\qquad\text{where }\dfrac{dx}{dt}\ne0.$

Proof. Let $t$ change by a small $\delta t\ne0$, producing corresponding changes $\delta x$ and $\delta y$ in $x$ and $y$ (with $\delta x\ne0$ since $\dfrac{dx}{dt}\ne0$). The incrementary ratio can be split exactly as $\dfrac{\delta y}{\delta x}=\dfrac{\delta y/\delta t}{\delta x/\delta t}.$ As $\delta t\to0$, both $\delta x\to0$ and $\delta y\to0$ (by continuity), and the right side tends to $\dfrac{dy/dt}{dx/dt}$, which exists and is finite because $x$ and $y$ are both differentiable in $t$. Since the right side has a finite limit, the left side $\displaystyle\lim_{\delta x\to0}\dfrac{\delta y}{\delta x}=\dfrac{dy}{dx}$ exists too, proving $y$ is differentiable in $x$, with $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}.$

Worked Example 1 — find $\dfrac{dy}{dx}$:

1. $x=at^4,\ y=2at^2$: $\dfrac{dy}{dt}=4at$, $\dfrac{dx}{dt}=4at^3$, so $\dfrac{dy}{dx}=\dfrac1{t^2}$.
2. $x=t-\sqrt t,\ y=t+\sqrt t$: $\dfrac{dy}{dt}=1+\dfrac1{2\sqrt t}=\dfrac{2\sqrt t+1}{2\sqrt t}$, $\dfrac{dx}{dt}=1-\dfrac1{2\sqrt t}=\dfrac{2\sqrt t-1}{2\sqrt t}$, so $\dfrac{dy}{dx}=\dfrac{2\sqrt t+1}{2\sqrt t-1}$.
3. $x=\cos(\log t),\ y=\log(\cos t)$: $\dfrac{dy}{dt}=-\tan t$, $\dfrac{dx}{dt}=-\dfrac{\sin(\log t)}{t}$, so $\dfrac{dy}{dx}=\dfrac{t\tan t}{\sin(\log t)}$.
4. $x=a(\theta+\sin\theta),\ y=a(1-\cos\theta)$: $\dfrac{dy}{d\theta}=a\sin\theta$, $\dfrac{dx}{d\theta}=a(1+\cos\theta)$, so $\dfrac{dy}{dx}=\dfrac{\sin\theta}{1+\cos\theta}=\tan\dfrac\theta2$ (using the half-angle identity).
5. $x=\sqrt{1-t^2},\ y=\sin^{-1}t$: $\dfrac{dy}{dt}=\dfrac1{\sqrt{1-t^2}}$, $\dfrac{dx}{dt}=-\dfrac{t}{\sqrt{1-t^2}}$, so $\dfrac{dy}{dx}=-\dfrac1t$. Worked Example 2 — evaluate $\dfrac{dy}{dx}$ at the given parameter value:

(i) $x=\sec^2\theta,\ y=\tan^3\theta$, at $\theta=\frac\pi3$: $\dfrac{dy}{dx}=\dfrac{3\tan^2\theta\sec^2\theta}{2\sec^2\theta\tan\theta}=\dfrac32\tan\theta$, so at $\theta=\frac\pi3$, $\dfrac{dy}{dx}=\dfrac{3\sqrt3}2$.

(ii) $x=t+\frac1t,\ y=\frac1{t^2}$, at $t=\frac12$: $\dfrac{dy}{dx}=\dfrac{-2/t^3}{1-1/t^2}=-\dfrac2{t(t^2-1)}$, so at $t=\frac12$, $\dfrac{dy}{dx}=\dfrac{16}3$.

(iii) $x=3\cos t-2\cos^3t,\ y=3\sin t-2\sin^3t$, at $t=\frac\pi6$: differentiating each using the identity $3\sin t-4\sin^3t=\sin3t$-type simplification gives $\dfrac{dy}{dt}=3\cos t\cos2t$ and $\dfrac{dx}{dt}=3\sin t\cos2t$, so $\dfrac{dy}{dx}=-\cot t$; at $t=\frac\pi6$, $\dfrac{dy}{dx}=-\cot\frac\pi6=-\sqrt3$.

Worked Example 3. If $x^2+y^2=t+\frac1t$ and $x^4+y^4=t^2+\frac1{t^2}$, show $x^3y\dfrac{dy}{dx}=-1$. Squaring the first relation: $(x^2+y^2)^2=t^2+2+\frac1{t^2}=(x^4+y^4)+2$, so $2x^2y^2=2$, i.e. $x^2y^2=1$. Differentiating this implicitly: $2x^2y\dfrac{dy}{dx}+2xy^2=0$, so $\dfrac{dy}{dx}=-\dfrac{y}{x}\cdot\dfrac1{x^2y^2}=-\dfrac1{x^3y}$ (using $x^2y^2=1$), i.e. $x^3y\dfrac{dy}{dx}=-1$. …