Q.If $y = (\cos^{-1}x)^2$, prove that $(1-x^2)\dfrac{d^2y}{dx^2} - x\dfrac{dy}{dx} - 2 = 0$. Hence find $y_2$ when $x=0$.

Parametric differentiation. When $x$ and $y$ are both given as functions of a third variable $t$ (the parameter), $x=f(t),\ y=g(t)$, the relationship between $x$ and $y$ is described parametrically rather than by a single equation. The chain rule gives

and, symmetrically, $\dfrac{dx}{dy}=\dfrac{f'(t)}{g'(t)}$. Worked example: the circle $x=r\cos t,\,y=r\sin t$ gives $\dfrac{dy}{dx}=\dfrac{r\cos t}{-r\sin t}=-\cot t$ — no elimination of $t$ is needed to find the tangent slope.

Differentiation of one function with respect to another. Given two functions $f(x)$ and $g(x)$ of the same variable $x$ with $g'(x)\ne0$, the derivative of $f$ with respect to $g$ is

— the same ratio idea as the parametric formula, with $x$ itself playing the role of the connecting parameter. When $g(x)=x$, this reduces to the ordinary derivative $f'(x)$.

Higher-order derivatives. If $f'$ is itself differentiable, its derivative is the second derivative, $f''(x)=\dfrac{d}{dx}\big[f'(x)\big]=\dfrac{d^2y}{dx^2}$; differentiating again gives the third derivative $f'''(x)=\dfrac{d^3y}{dx^3}$, and so on. Physically, if $s=s(t)$ is position, $v=s'$ is velocity, $a=v'=s''$ is acceleration, and $j=a'=s'''$ is jerk (the rate of change of acceleration). Geometrically, the first derivative is a slope; the second derivative measures a rate of change of that slope (its precise geometric meaning — curvature — is developed in later study). …