Math for Non-Geeks/ Derivatives of higher order

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Motivation

The derivative $f^{\prime }$ describes the current rate of change of the function $f$. Now the derivative function $f^{\prime }$ can be differentiated again, provided that it is again differentiable. The obtained derivative of the derivative is called second derivative or derivative of second order and is called $f^{″}$ or $f^{(2)}$. This can be done arbitrarily often. If the second derivative is again differentiable, a third derivative $f^{(3)}$ can be constructed, then a fourth derivative $f^{(4)}$ and so on.

These higher derivatives allow statements about the course of a function graph. The second derivative tells us whether a graph is curved upwards ("convex") or curved downwards ("concave"). If a function has a convex graph, its gradient increases continuously. For this convexity, $f^{″}(x)>0$ is a sufficient condition. If the second derivative is always positive, then the first derivative must grow continuously. Analogously, it follows from $f^{″}(x)<0$ that the graph is concave and the derivative falls monotonically.

Higher-order derivatives do not only tell us more about abstract functions, they can also have a physical meaning. Consider the function $f:[a,b]\to ℝ$ with $f(t)=t^3+2$, which shall describe the location $f(t)$ of a car at the time $t$. We already know that we can calculate the speed of the car at the time $t$ with the first derivative: $f^{\prime }(t)=f^{(1)}(t)=3t^2$. What does the derivative $f^{″}(t)$ of $f^{\prime }$ say? This is the instantaneous rate of change of speed and thus the acceleration of the car. It accelerates with $f^{″}(t)=f^{(2)}(t)=6t$. So second derivatives describe accelerations.

Now we can derive this second derivative again, whereby we get the rate of change of acceleration $f^{‴}(t)=f^{(3)}(t)=6$. This is called jerk in vehicle dynamics and indicates how fast a car increases acceleration or how fast it initiates braking. For example, a big jerk occurs during emergency braking. Since $f^{(3)}(t)<0$ is in an emergency stop, the graph of the speed $f^{\prime }$ is convex - the speed decreases more and more. The fourth derivative $f^4(t)=0$ again tells us that the jerk has no instantaneous rate of change.

Definition

The set of all $k$ times continuously differential functions with domain of definition $D$ and range $W$ is denoted $C^k(D,W)$. In particular $C^0(D,W)=C(D,W)$ consists of the continuous functions. If we can derive the function $f$ arbitrarily often, we write $f\in C^{\mathrm{\infty }}(D,W)$. If $W=ℝ$, then we can write $C^k(D)$ or $C^{\mathrm{\infty }}(D)$ in short. Those sets of functions satisfy the inclusion chain:

Computation rules for higher derivatives

Linearity

The linearity of derivatives is also "inherited" to higher derivatives: If $f$ and $g$ are differentiable, for $a,b\in ℝ$ the function $af+bg$ is also differentiable with

If $f$ and $g$ are now even twice differentiable, then there is

If we continue to do so, we will get

Leibniz rule for product functions Template:Anchor

We now try to determine a general formula for the $n$-th derivative of the product function $fg$ of two arbitrarily often differentiable functions $f$ and $g$. By applying the factor-, sum- and product rule several times we obtain for $n=1,2,3$

If we plug in $f=x$ and $g=y$, and instead of the derivatives of $f$ and $g$ the corresponding powers of $x$ and $y$, we see a clear analogy to the binomial theorem:

This analogy can be made clear as follows:

We assign for every $k\in {\mathbb{N}}_0$ the derivative $f^{(k)}$ to the power $x^k$, and the derivative $g^{(k)}$ to the power $y^k$. The $0$-th derivative $f^{(0)}=f$ corresponds to the $0$-th power $x^0=1$. The derivative of the term $f^{(k)}{g}^{(l)}$ is by means of the product rule

The expression $f^{(k+1)}{g}^{(l)}+f^{(k)}{g}^{(l+1)}$ now corresponds in our analogy to the sum $x^{k+1}{y}^l+x^k{y}^{l+1}$. We get this term from $x^k{y}^l$ by multiplication with $x+y$. For our polynomials, the distributive law yields

Therefore, the application of the product rule corresponds to the multiplication with the sum $x+y$. Thus the $n$-th derivative $(fg{)}^{(n)}$ corresponds to the power $(x+y{)}^n$. From the binomial theorem

we hence get the

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