Parametric, Implicit & Higher-Order Derivatives

Differentiate $F(x,y)=0$ term by term w.r.t. $x$: a term in $y$ gives its derivative $\times \frac{dy}{dx}$ (chain rule), and products use the product rule. Collect the $\frac{dy}{dx}$ terms and solve. No need to express $y$ explicitly.

For relations like $x^y y^x = c$: take $\ln$ to get $y\ln x + x\ln y = \ln c$, then differentiate implicitly (product rule on each term, $\frac{dy}{dx}$ on $y$-terms) and solve. For $x^m y^n = k$: $\,m\ln x + n\ln y = \ln k \Rightarrow \frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = 0$.

If $x=f(t)$ and $y=g(t)$, then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{g'(t)}{f'(t)}$ (provided $f'(t)\neq 0$). The result is usually left in terms of $t$. Second derivative: differentiate $\dfrac{dy}{dx}$ (a function of $t$) w.r.t. $t$ and divide by $\dfrac{dx}{dt}$ again: $\dfrac{d^2y}{dx^2} = \dfrac{\frac{d}{dt}\big(\frac{dy}{dx}\big)}{dx/dt}$. It is NOT $\dfrac{d^2y/dt^2}{d^2x/dt^2}$. For $x=t^2,\ y=t^3$: $\dfrac{dy}{dx}=\tfrac{3t}{2}$, so $\dfrac{d^2y}{dx^2} = \dfrac{3/2}{2t} = \dfrac{3}{4t}$.

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right)$, and so on. Useful standard results: $(e^{ax})^{(n)} = a^n e^{ax}$; $(\sin x)^{(n)} = \sin\!\left(x + \tfrac{n\pi}{2}\right)$. To evaluate at a point, differentiate first and substitute the value at the end.

• Kinematics reading: for position $s(t)$, velocity $v=\dfrac{ds}{dt}$ and acceleration $a=\dfrac{d^2s}{dt^2}=\dfrac{dv}{dt}$; going back, $v=\int a\,dt$ and $s=\int v\,dt$ with the constants fixed by initial values.
• Sign of a combination like $f+f'+f''$: for a quadratic $f$, $g=f+f'+f''$ is again a quadratic; it is positive for ALL real $x$ iff its leading coefficient is positive AND its discriminant is negative ($b^2-4ac<0$) — both conditions, not one.

$\dfrac{dx}{dy} = \left(\dfrac{dy}{dx}\right)^{-1}$, and differentiating this w.r.t. $y$ (chain rule) gives $\dfrac{d^2x}{dy^2} = -\dfrac{d^2y/dx^2}{\left(dy/dx\right)^{3}}$. Memorise the cube in the denominator and the minus sign — the classic trap.

Differentiate $y$ once and twice, then substitute into the target relation and simplify — the goal is to eliminate $\sin, \ln,$ etc. and land on $0$. Example: $y=\sin(\ln x)$ gives $xy' = \cos(\ln x)$, and differentiating again yields $x^2 y'' + x y' + y = 0$.