Higher order Derivatives

If $s=s(t)$ is the position (displacement) of an object moving along a straight line, its first derivative already has a direct physical meaning: the velocity $v(t) = s'(t) = \dfrac{ds}{dt} = \lim_{\Delta t\to0}\dfrac{s(t+\Delta t)-s(t)}{\Delta t}$ — this is exactly the instantaneous velocity defined in §10.2.2. The instantaneous rate of change of velocity with respect to time is called the acceleration $a(t)$ of the object; being the derivative of $v(t)$, it is therefore the second derivative of the position function:

More generally, for any differentiable function $f$, the first derivative $f'(x)$ is itself a function of $x$ and so may itself have a derivative — if it exists, it is denoted $f''=(f')'$ and called the second derivative:

Other equivalent notations: $D^2f(x)$, $D^2y$, $y''$. Geometrically, while $f'$ has the simple reading "slope of the tangent", the second derivative measures a rate of change of a rate of change; its geometric meaning is subtler (it connects, in later study, to the radius of curvature of the graph), but the physical reading — acceleration, when $y$ is position — is immediate.

Higher orders still. If $f''$ itself is differentiable, its derivative is the third derivative, $f'''(x) = \dfrac{d^3y}{dx^3} = y'''$. Physically, when $y=s(t)$ is position, $s'''(t) = a'(t)$ is called the jerk: $j = \dfrac{da}{dt} = \dfrac{d^3s}{dt^3}$ — the rate of change of acceleration, aptly named because a large jerk means a sudden change in acceleration, producing an abrupt jolt.

Worked illustration — explicit polynomial. For $y=x^3-6x^2-5x+3$: $y'=3x^2-12x-5$, then $y''=6x-12$, then $y'''=6$ (a constant — differentiating a cubic three times always eventually gives a constant, and a fourth derivative would be $0$).

Worked illustration — a power with a negative exponent. For $y=1/x=x^{-1}$: $y'=-x^{-2}$, $y''=\dfrac{(-1)(-2)}{x^3}=\dfrac{2!}{x^3}$, and $y'''=\dfrac{(-1)(-2)(-3)}{x^4}=-\dfrac{3!}{x^4}$ — each further derivative of $x^{-1}$ brings down one more factorial-growing constant with alternating sign.

Worked illustration — a product, needing product + chain rule twice. For $f(x)=x\cos x$: $f'(x) = -x\sin x+\cos x$ (product rule); differentiating again (product rule on $-x\sin x$, plus the derivative of $\cos x$), $f''(x) = -(x\cos x+\sin x)-\sin x = -x\cos x-2\sin x$.

Worked illustration — implicit second derivative. For $x^4+y^4=16$: differentiating implicitly, $4x^3+4y^3y'=0$, so $y'=-x^3/y^3$. To find $y''$, differentiate this expression for $y'$ using the quotient rule, remembering $y$ is itself a function of $x$: $y'' = \dfrac{y^3(-3x^2)-(-x^3)(3y^2y')}{y^6} = \dfrac{-3x^2y^3+3x^3y^2y'}{y^6}$; substituting $y'=-x^3/y^3$ and simplifying yields $y'' = -\dfrac{3x^2}{y^7}\big(y^4+x^4\big) = -\dfrac{3x^2(16)}{y^7} = -\dfrac{48x^2}{y^7}$ (using the original equation $x^4+y^4=16$ to simplify). …