Second Order Derivatives

Definition. If $y=f(x)$ is differentiable, its derivative $dy/dx=f'(x)$ is itself a function

of $x$; if that function is in turn differentiable, its derivative is called the second order derivative (or second derivative) of $y$ with respect to $x$, written

Physically, if $y=s(t)$ is the position of a particle at time $t$, then $dy/dt$ is its velocity

and $d^2y/dt^2$ is its acceleration -- the rate of change of the rate of change; more

generally, $d^2y/dx^2$ measures how the slope of the curve $y=f(x)$ itself is changing.

For an explicit function. If $y=f(x)$ is given explicitly, finding $y''$ is simply a matter

of differentiating twice in succession: find $y'=dy/dx$ first, then differentiate that result

again with respect to $x$.

For an implicitly-defined function. If $y$ is defined implicitly by an equation (Section 5),

the first derivative $dy/dx$ typically comes out as an expression in both $x$ and $y$; to find

$d^2y/dx^2$, differentiate this first-derivative expression again with respect to $x$, using the

product/quotient rule as needed and substituting $dy/dx$ (already found) wherever a $y$-term is

differentiated. For $x^2+y^2=25$ (Example 5), $dy/dx=-x/y$, so

using $x^2+y^2=25$ in the last step to simplify the numerator back to a constant.

For a parametric function -- the one genuine trap. If $x=f(t),\ y=g(t)$, the first derivative

is $dy/dx=(dy/dt)/(dx/dt)$ (Section 9), itself a function of $t$. The second derivative is

not simply $\dfrac{d^2y/dt^2}{d^2x/dt^2}$ -- that shortcut is a common and serious error.

Instead, treat $dy/dx$ as a new function of $t$ and apply the parametric-derivative idea a second

time:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}.$ …