Derivatives of higher order – "Math for Non-Geeks"

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

Definition (Derivatives of higher order)

Let $f:D\to W$ with $D,W\subseteq ℝ$ be a real function. We set $f^{(0)}:=f$ and in the case of differentiability $f^{(1)}=f^{\prime }$. We define the second derivative via $f^{(2)}={\left(f^{(1)}\right)}^{\prime }$, the third derivative via $f^{(3)}={\left(f^{(2)}\right)}^{\prime }$ etc., if these higher derivatives exist. Overall, we define recursively for $k\in {\mathbb{N}}_0$:

$f^{(k+1)}:={\left(f^{(k)}\right)}^{\prime }$

We say $f$ that is $k$ times differentiable, if the $k$-th derivative $f^{(k)}$ of $f$ exists. $f$ is called $k$ times continuously differentiable, if $f^{(k)}$ is continuous (which is a stronger statement).

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:

$C(D,W)\supseteq C^1(D,W)\supseteq C^2(D,W)\supseteq \dots \supseteq C^k(D,W)\supseteq C^{k+1}(D,W)$

Question: Are the following statements true of false?

1. $|x|\in C(ℝ)$
2. $|x|\in C^1(ℝ)$
3. $\mathrm{sgn}(x)\in C(ℝ)$
4. $\mathrm{sgn}(x)\in C^1(ℝ)$
5. $x\in C(ℝ)$
6. $x\in C^1(ℝ)$

Solutions:

1. true
2. false
3. false
4. false
5. true
6. true

Examples for higher derivatives

Derivatives of the power function

Example (Derivatives of the power function)

We consider the function $f:ℝ\to ℝ,x\mapsto x^2$. This function is infinitely often differentiable, since there is for all $x\in ℝ$ and all $k\in \mathbb{N},k\ge 3$:

$\begin{array}{cc}& f^{\prime }(x)=2x\\ & f^{″}(x)=2\\ & f^{(k)}=0\\ \end{array}$

In general, for $f:ℝ\to ℝ,x\mapsto x^n$ with $n\in \mathbb{N}$ there is:

$\begin{array}{cc}& f^{\prime }(x)=n{x}^{n-1}\\ & f^{(k)}(x)=n(n-1)\cdot \dots \cdot (n-k+1)x^{n-k}\text{ for }k\le n\\ & f^{(k)}=0\text{ for }k>n\\ \end{array}$

Derivatives of the exponential function

Example (Derivatives of the exponential function)

For the exponential function $\exp :ℝ\to ℝ$ since ${\exp }^{\prime }(x)=\exp (x)$ for all $x\in ℝ$ we have infinite differentiability $\exp \in C^{\mathrm{\infty }}(ℝ,ℝ)$. In addition, there is for all $k\in \mathbb{N}$:

${\exp }^{(k)}(x)=\exp (x)$

Derivatives of the sine function

Example (Derivatives of the sine function)

The function $\sin :ℝ\to ℝ$ is infinitely often continuously differentiable. For all $x\in ℝ$ there is:

$\begin{array}{cc}& {\sin }^{\prime }(x)=\cos (x)\\ & {\sin }^{″}(x)={\cos }^{\prime }(x)=-\sin (x)\\ & {\sin }^{‴}(x)=(-\sin {)}^{\prime }(x)=-\cos (x)\\ & {\sin }^{(4)}(x)=(-\cos {)}^{\prime }(x)=\sin (x)\\ \end{array}$

In general, for all $n\in {\mathbb{N}}_0$ there is:

$\begin{array}{cc}& {\sin }^{(4n+1)}(x)=\cos (x)\\ & {\sin }^{(4n+2)}(x)={\cos }^{\prime }(x)=-\sin (x)\\ & {\sin }^{(4n+3)}(x)=(-\sin {)}^{\prime }(x)=-\cos (x)\\ & {\sin }^{(4n)}(x)=(-\cos {)}^{\prime }(x)=\sin (x)\\ \end{array}$

Question: What are the derivatives of $\cos :ℝ\to ℝ$?

We use that ${\sin }^{\prime }(x)=\cos (x)$. For $x\in ℝ$ there is

$\begin{array}{cc}& {\cos }^{\prime }(x)=-\sin (x)\\ & {\cos }^{″}(x)=-\cos (x)\\ & {\cos }^{‴}(x)=\sin (x)\\ & {\cos }^{(4)}(x)=\cos (x)\end{array}$

In general, for all $n\in {\mathbb{N}}_0$ there is:

$\begin{array}{cc}& {\cos }^{(4n+1)}(x)=-\sin (x)\\ & {\cos }^{(4n+2)}(x)=-\cos (x)\\ & {\cos }^{(4n+3)}(x)=\sin (x)\\ & {\cos }^{(4n)}(x)=\cos (x)\end{array}$

Exercises: higher derivatives

Derivatives of the logarithm function

Exercise (Derivatives of the logarithm function)

Show that the logarithm function $\ln :{ℝ}^{+}\to ℝ$ is arbitrarily often differentiable and that for all $n\in \mathbb{N}$ there is:

${\ln }^{(n)}(x)=(-1{)}^{n-1}\frac{(n-1)!}{x^n}$

Proof (Derivatives of the logarithm function)

Theorem whose validity shall be proven for $n\in \mathbb{N}$ bewiesen werden soll:

${\ln }^{(n)}(x)=(-1{)}^{n-1}\frac{(n-1)!}{x^n}$

1. Base case:

${\ln }^{\prime }(x)=\frac{1}{x}=(-1{)}^0\frac{0!}{x^1}$

2. Inductive step:

2a. Inductive hypothesis:

${\ln }^{(n)}(x)=(-1{)}^{n-1}\frac{(n-1)!}{x^n}$

2b. Induction theorem:

${\ln }^{(n+1)}(x)=(-1{)}^n\frac{(n)!}{x^{n+1}}$

2c. Proof of induction step:

$\begin{array}{cc}{\ln }^{(n+1)}(x)& =({\ln }^{(n)}(x){)}^{\prime }\underset{\text{assumption}}{\stackrel{\text{induction}}{=}}{((-1{)}^{n-1}\frac{(n-1)!}{x^n})}^{\prime }\\ & ={((-1{)}^{n-1}(n-1)!x^{-n})}^{\prime }\underset{\text{rules}}{\stackrel{\text{derivative}}{=}}(-1{)}^{n-1}(n-1)!(-n)x^{-n-1}\\ & =(-1{)}^n\frac{n!}{x^{n+1}}\end{array}$

Exactly once differentiable function

Exercise (Exactly once differentiable function)

Prove that the following function is differentiable once, but not twice:

$f:ℝ\to ℝ,x\mapsto \{\begin{array}{cc}{x}^2\sin \left(\frac{1}{x}\right)& ,x\ne 0\\ 0& ,x=0\end{array}$

Solution (Exactly once differentiable function)

The function $f$ with $f(x)=x^2\sin \left(\frac{1}{x}\right)$ for $x\ne 0$ and $f(0)=0$.

This function is differentiable in all points $\tilde{x}\ne 0$, since for all $x$ in the open neighbourhood $(0,2\tilde{x})$ for $\tilde{x}>0$ or $(2\tilde{x},0)$ for $\tilde{x}<0$ there is $f(x)=x^2\sin \left(\frac{1}{x}\right)$ . Consequently, by the product and the chain rule

$f^{\prime }(\tilde{x})=2\tilde{x}\cdot \sin \left(\frac{1}{\tilde{x}}\right)+{\tilde{x}}^2\cos \left(\frac{1}{\tilde{x}}\right)\cdot \frac{-1}{{\tilde{x}}^2}=2\tilde{x}\cdot \sin \left(\frac{1}{\tilde{x}}\right)-\cos \left(\frac{1}{\tilde{x}}\right)$

For $\tilde{x}=0$ we obtain

$\begin{array}{cc}& \underset{x\to 0}{lim}\frac{f(x)-f(0)}{x-0}\\ & =\underset{x\to 0}{lim}\frac{x^2\sin \left(\frac{1}{x}\right)-0}{x}\\ & =\underset{x\to 0}{lim}x\sin \left(\frac{1}{x}\right)\\ & =0\end{array}$

Since for all $x\in ℝ\setminus \{0\}$ there is $\sin \left(\frac{1}{x}\right)\in (-1,1)$ , so the term is bounded. Hence, for the derivative function

$f^{\prime }:ℝ\to ℝ,x\mapsto \{\begin{array}{cc}2x\cdot \sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)& ,x\ne 0\\ 0& ,x=0\end{array}$

However, this function is not differentiable at $\tilde{x}=0$. We approach 0 by taking two sequences $(a_n{)}_{n\in \mathbb{N}}$ and $(b_n{)}_{n\in \mathbb{N}}$, where we define for all $n\in \mathbb{N}$

$\begin{array}{cc}& a_n=\frac{1}{\frac{\pi }{2}+2\pi n}\\ & b_n=\frac{1}{\frac{3\pi }{2}+2\pi n}\\ \end{array}$

Then, there is $\underset{n\to \mathrm{\infty }}{lim}{a}_n=0$ and $\underset{n\to \mathrm{\infty }}{lim}{b}_n=0$. Further there is for all $n\in \mathbb{N}$

$\begin{array}{cc}& f^{\prime }(a_n)=2\frac{1}{\frac{\pi }{2}+2\pi n}\cdot \sin (\frac{\pi }{2}+2\pi n)-\cos (\frac{\pi }{2}+2\pi n)=\frac{2}{\frac{\pi }{2}+2\pi n}\\ & f^{\prime }(b_n)=2\frac{1}{\frac{3\pi }{2}+2\pi n}\cdot \sin (\frac{3\pi }{2}+2\pi n)-\cos (\frac{3\pi }{2}+2\pi n)=\frac{-2}{\frac{\pi }{2}+2\pi n}\\ \end{array}$

So there is

$\begin{array}{cc}& \underset{n\to \mathrm{\infty }}{lim}\frac{f^{\prime }(a_n)-f^{\prime }(0)}{a_n-0}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{2}{\frac{\pi }{2}+2\pi n}-0)}{\frac{1}{\frac{\pi }{2}+2\pi n}-0}=2\end{array}$

But

$\begin{array}{cc}& \underset{n\to \mathrm{\infty }}{lim}\frac{f^{\prime }(b_n)-f^{\prime }(0)}{b_n-0}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{-2}{\frac{3\pi }{2}+2\pi n}-0}{\frac{1}{\frac{3\pi }{2}+2\pi n}-0}=-2\end{array}$

Consequently, the limit value $\underset{x\to 0}{lim}\frac{f^{\prime }(x)-f^{\prime }(0)}{x-0}$ does not exist and therefore $f^{\prime }$ is not differentiable at $0$ .

Additional question: Is $f^{\prime }$ continuous at $0$?

Nope. Take the two sequences:

$\begin{array}{cc}& (a_n{)}_{n\in \mathbb{N}}=(\frac{1}{2n\pi }{)}_{n\in \mathbb{N}}\\ & (b_n{)}_{n\in \mathbb{N}}=(\frac{1}{(2n+1)\pi }{)}_{n\in \mathbb{N}}\\ \end{array}$

For these sequences, there is: $\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{b}_n=0$. However

$\begin{array}{cc}& f^{\prime }(a_n)=2\frac{1}{2n\pi }\cdot \underset{=0}{\underbrace{\sin \left(2n\pi \right)}}-\underset{=1}{\underbrace{\cos \left(2n\pi \right)}}=-1\\ & f^{\prime }(b_n)=2\frac{1}{(2n+1)\pi }\cdot \underset{=0}{\underbrace{\sin ((2n+1)\pi )}}-\underset{=-1}{\underbrace{\cos ((2n+1)\pi )}}=1\end{array}$

So $\underset{x\to 0}{lim}{f}^{\prime }(x)$ doesn't exist. By means of the sequence criterion, $f^{\prime }$ is hence not continuous at $0$.

Remark: Therefore, $f^{\prime }$ is also not differentiable at $0$.

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

$(af+bg{)}^{\prime }=a{f}^{\prime }+b{g}^{\prime }$

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

$(af+bg{)}^{″}=(a{f}^{\prime }+b{g}^{\prime }{)}^{\prime }=a{f}^{″}+b{g}^{″}$

If we continue to do so, we will get

Theorem (Linearity of higher derivatives)

Let $a,b\in ℝ$ and $f,g:D\to ℝ$ be $n$ times differentiable. Then also $af+bg:D\to ℝ$ is $n$ times differentiable, and for all $n\in {\mathbb{N}}_0$ there is:

$(af+bg{)}^{(n)}=a{f}^{(n)}+b{g}^{(n)}$

Example (Linearity of higher derivatives)

Since ${\sin }^{(4n+1)}=\cos$ and ${\cos }^{(4n+1)}=-\sin$ for $n\in \mathbb{N}$ there is

$(3\sin (x)+4\cos (x){)}^{(1001)}=3{\sin }^{(1001)}(x)+4{\cos }^{(1001)}(x)=3{\sin }^{(4\cdot 250+1)}(x)+4{\cos }^{(4\cdot 250+1)}(x)=3\cos (x)-4\sin (x)$

Proof (Linearity of higher derivatives)

Theorem whose validity shall be proven for $n\in \mathbb{N}$ bewiesen werden soll:

$(af+bg{)}^{(n)}=a{f}^{(n)}+b{g}^{(n)}$

1. Base case:

$(af+bg{)}^{(0)}=af+bg=a{f}^{(0)}+b{g}^{(0)}$

2. Inductive step:

2a. Inductive hypothesis:

$(af+bg{)}^{(n)}=a{f}^{(n)}+b{g}^{(n)}$

2b. Induction theorem:

$(af+bg{)}^{(n+1)}=a{f}^{(n+1)}+b{g}^{(n+1)}$

2c. Proof of induction step:

$\begin{array}{cc}(af+bg{)}^{(n+1)}& =((af+bg{)}^{(n)}{)}^{\prime }\\ & \downarrow \text{induction assumption}\\ & =(a{f}^{(n)}+b{g}^{(n)}{)}^{\prime }\\ & \downarrow \text{linearity of the derivative}\\ & =a{f}^{(n+1)}+b{g}^{(n+1)}\end{array}$

Leibniz rule for product functions

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 $n=1,2,3$

$\begin{array}{cc}(fg{)}^{\prime }& =f^{\prime }g+f{g}^{\prime }\\ (fg{)}^{″}& =(f^{\prime }g+g^{\prime }f{)}^{\prime }=(f^{\prime }g{)}^{\prime }+(f{g}^{\prime }{)}^{\prime }\\ & =f^{″}g+f^{\prime }{g}^{\prime }+f^{\prime }{g}^{\prime }+f{g}^{″}\\ & =f^{″}g+2f^{\prime }{g}^{\prime }+f{g}^{″}\\ (fg{)}^{‴}& =(f^{″}g+2f^{\prime }{g}^{\prime }+f{g}^{″}{)}^{\prime }\\ & =(f^{″}g{)}^{\prime }+(2f^{\prime }{g}^{\prime }{)}^{\prime }+(f{g}^{″}{)}^{\prime }\\ & =f^{‴}g+f^{″}{g}^{\prime }+2f^{″}{g}^{\prime }+2f^{\prime }{g}^{″}+f^{\prime }{g}^{″}+f{g}^{‴}\\ & =f^{‴}g+3f^{″}{g}^{\prime }+3f^{\prime }{g}^{″}+f{g}^{‴}\end{array}$

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:

$\begin{array}{cc}(x+y{)}^1& =x^1{y}^0+x^0+y^1\\ (x+y{)}^2& =x^2+2xy+y^2\\ & =x^2{y}^0+2x^1{y}^1+x^0{y}^2\\ (x+y{)}^3& =x^3+3x^{2}y+3x{y}^2+y^3\\ & =x^3{y}^0+3x^2{y}^1+3x^1{y}^2+3x^0{y}^3\end{array}$

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

$(f^{(k)}{g}^{(l)}{)}^{\prime }=f^{(k+1)}{g}^{(l)}+f^{(k)}{g}^{(l+1)}$

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

$(x^k{y}^l)(x+y)=x^{k+1}{y}^l+x^k{y}^{l+1}$

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

$(x+y{)}^n=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})x^k{y}^{n-k}$

we hence get the

Theorem (Leibniz rule for derivatives)

Let $f,g:D\to ℝ$ be $n$ times differentiable functions. Then, $fg:D\to ℝ$ is $n$ times differentiable, and for all $n\in {\mathbb{N}}_0$, there is:

$(fg{)}^{(n)}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k)}$

Example (Leibniz rule for derivatives)

Using the Leibniz rule we calculate $(x^3{e}^x{)}^{(2016)}$. The rule is applicable because $x\mapsto x^3$ and $x\mapsto e^x$ are arbitrarily often differentiable on $ℝ$. There is

$\begin{array}{cc}(x^3{e}^x{)}^{(2016)}& \underset{\text{rule}}{\stackrel{\text{Leibniz}}{=}}\sum _{k=0}^{2016}(\genfrac{}{}{0ex}{}{2016}{k})(x^3{)}^{(k)}(e^x{)}^{(2016-k)}\\ & \downarrow (x^3{)}^{\prime }=3x^2,(x^3{)}^{″}=6x,(x^3{)}^{(3)}=6,(x^3{)}^{(k)}=0\text{ for }k\ge 4,(e^x{)}^{(k)}=e^x\text{ for all }k\in \mathbb{N}\\ & =(\genfrac{}{}{0ex}{}{2016}{0})x^3{e}^x+(\genfrac{}{}{0ex}{}{2016}{1})\cdot 3x^2{e}^x+(\genfrac{}{}{0ex}{}{2016}{2})\cdot 6x{e}^x+(\genfrac{}{}{0ex}{}{2016}{3})\cdot 6e^x\\ & =x^3{e}^x+2016\cdot 3x^2{e}^x+\frac{2016\cdot 2015}{2}\cdot 6x{e}^x+\frac{2016\cdot 2015\cdot 2014}{3\cdot 2}\cdot 6e^x\\ & =x^3{e}^x+2016\cdot 3x^2{e}^x+2016\cdot 2015\cdot 3x{e}^x+2016\cdot 2015\cdot 2014e^x\end{array}$

Proof (Leibniz rule for derivatives)

Theorem whose validity shall be proven for $n\in \mathbb{N}$ bewiesen werden soll:

$(fg{)}^{(n)}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k)}$

1. Base case:

$(fg{)}^{(0)}=fg=\sum _{k=0}^0(\genfrac{}{}{0ex}{}{0}{k})f^{(k)}{g}^{(0-k)}$

2. Inductive step:

2a. Inductive hypothesis:

$(fg{)}^{(n)}=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k)}$

2b. Induction theorem:

$(fg{)}^{(n+1)}=\sum _{k=0}^{n+1}(\genfrac{}{}{0ex}{}{n+1}{k})f^{(k)}{g}^{(n+1-k)}$

2c. Proof of induction step:

$\begin{array}{cc}(fg{)}^{(n+1)}& ={[(fg{)}^{(n)}]}^{\prime }\\ & \downarrow \text{ induction assumption}\\ & ={\left[\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k)}\right]}^{\prime }\\ & \downarrow \text{ linearity of the derivative}\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k}){\left[f^{(k)}{g}^{(n-k)}\right]}^{\prime }\\ & \downarrow \text{ product rule}\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})[f^{(k+1)}{g}^{(n-k)}+f^{(k)}{g}^{(n-k+1)}]\\ & =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k+1)}{g}^{(n-k)}+\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k+1)}=\\ & \downarrow \text{ index shift}\\ & =\sum _{k=1}^{n+1}(\genfrac{}{}{0ex}{}{n}{k-1})f^{(k)}{g}^{(n-(k-1))}+\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k+1)}\\ & \downarrow (\genfrac{}{}{0ex}{}{n}{-1})=0\text{ and }(\genfrac{}{}{0ex}{}{n}{n+1})=0\\ & =\sum _{k=0}^{n+1}(\genfrac{}{}{0ex}{}{n}{k-1})f^{(k)}{g}^{(n-k+1)}+\sum _{k=0}^{n+1}(\genfrac{}{}{0ex}{}{n}{k})f^{(k)}{g}^{(n-k+1)}\\ & =\sum _{k=0}^{n+1}[(\genfrac{}{}{0ex}{}{n}{k-1})+(\genfrac{}{}{0ex}{}{n}{k})]f^{(k)}{g}^{(n-k+1)}\\ & \downarrow \text{ sum of binomial coefficients}\\ & =\sum _{k=0}^{n+1}(\genfrac{}{}{0ex}{}{n+1}{k})f^{(k)}{g}^{((n+1)-k)}\end{array}$