Definition 2.2.1. Let $I$ be an open interval in $\mathbb R$. The differentiation operator $D$ is the operator that sends each differentiable function $f : I \to \mathbb R$ to its derivative $f’$. In other words, $Df$ is the function defined by
\[(Df)(x) = f'(x)\]for every $x$ in $I$.
Theorem 2.2.2. Let be $\mathbf c$ a constant function and $\mathrm{id}$ the identity function. Then
- $D\mathbf c=\mathbf 0$
- $D(\mathrm{id})=\mathbf 1$
Theorem 2.2.3. If $f,g$ are differentiable function and $\mathbf c$ is a constant function, then
- $D(f+g)=Df+Dg$
- $D(f-g)=Df-Dg$
- $D(\mathbf c f)=\mathbf c Df$
- $D(fg)=(Df)g+f(Dg)$
Theorem 2.2.4. If $f,g$ are differentiable function, and $g(x)\ne 0$, then
\[D\left(\frac{f}{g}\right)=\frac{(Df)g-f(Dg)}{g^2}.\]Theorem 2.2.5.
- $D(\sin)=\cos$
- $D(\cos)=-\sin$
- $D(\tan)=\sec^2$
Theorem 2.2.6 (The Chain Rule). Let $g$ be differentiable at $x$, and let $f$ be differentiable at $g(x)$. Then
\[D(f\circ g)=((Df)\circ g)(Dg).\]Proof. Let $k=f\circ g$. Then we have $D(f\circ g)=Dk$. Thus $(D(f\circ g))(x)=k’(x)$. Let $\Delta_h = g(x+h)-g(x)$ and let $\phi$ be the function such that
\[\phi(t)= \begin{cases} \dfrac{f(g(x)+t)-f(g(x))}{t}, & t\ne 0,\\ f'(g(x)), & t=0. \end{cases}\]Then we have
\[\begin{aligned} \frac{k(x+h)-k(x)}{h} &=\frac{f(g(x+h))-f(g(x))}{h}\\ &=\phi(\Delta_h) \times \frac{\Delta_h}{h}. \end{aligned}\]Let $\varepsilon>0$ be given. Since $\phi$ is continuous at $0$, there exists $\eta>0$ such that $0<\vert \Delta_h \vert < \eta \Rightarrow \left\vert \phi(\Delta_h)-\phi(0)\right\vert<\varepsilon.$ Since $g$ is continuous at $x$, there exists $\delta>0$ such that $0<\vert h \vert <\delta \Rightarrow \vert \Delta_h\vert < \eta$. Thus $0<\vert h \vert <\delta \Rightarrow \left\vert \phi(\Delta_h)-\phi(0)\right\vert<\varepsilon.$ By Definition 1.3.2, $\lim_{h \to 0}{\phi(\Delta_h)}=\phi(0)=f’(g(x)).$ Thus we have
\[\begin{aligned} k'(x) &=\lim_{h\to 0}{\left[\phi(\Delta_h) \times \frac{\Delta_h}{h}\right]}\\ &=\lim_{h\to 0}{\phi(\Delta_h)} \times \lim_{h\to 0}{\frac{g(x+h)-g(x)}{h}}\\ &=f'(g(x))\times g'(x). \end{aligned}\]Finally, $(D(f\circ g))(x)=f’(g(x))\times g’(x)=((Df)\circ g)(x)\times (Dg)(x)$. Therefore $D(f\circ g)=((Df)\circ g)(Dg)$. $\square$
Exercises
Exercise 2.2.1. Prove that $D(f+g)=Df+Dg$.
Proof.
Claim 1. $\text{LHS}=f’+g’$.
Proof of Claim 1. We have $(Df)(x)=f’(x) \Rightarrow (Df)(x)+(Dg)(x)=f’(x)+(Dg)(x)$. And, also $(Dg)(x)=g’(x) \Rightarrow (Dg)(x)+f’(x)=g’(x)+f’(x)$. Thus $(Df)(x)+(Dg)(x)=f’(x)+g’(x)$. Therefore $\text{LHS}=Df+Dg=f’+g’$.$\Diamond$
Claim 2. $\text{RHS}=f’+g’$.
Proof of Claim 2. Let $f+g=k$. We have $(D(f+g))(x)=k’(x)$ and, $k(x)=(f+g)(x)=f(x)+g(x)$. Since
if follows that
\[\begin{aligned} \lim_{h\to 0}\frac{k(x+h)-k(x)}{h} &=\lim_{h\to 0}{\left[\frac{f(x+h)-f(x)}{h} + \frac{g(x+h)-g(x)}{h}\right]}\\ &=\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}} + \lim_{h\to 0}{\frac{g(x+h)-g(x)}{h}}\\ &=f'(x)+g'(x). \end{aligned}\]Thus $(D(f+g))(x)=f’(x)+g’(x)$. Therefore $\text{RHS}=D(f+g)=f’+g’$.$\Diamond$
By Claim 1 and Claim 2, $\text{LHS}=\text{RHS}$.$\square$
Exercise 2.2.2. Prove that $D(fg)=(Df)g+f(Dg)$.
Proof. Let $fg=k$. Then we have $D(fg)=Dk=k’$, and $k(x)=f(x)g(x)$. It follows that
Thus
\[\begin{aligned} \lim_{h\to 0}\frac{k(x+h)-k(x)}{h} &=\lim_{h\to 0}{\left[\frac{f(x+h)-f(x)}{h}g(x+h)+\frac{g(x+h)-g(x)}{h}f(x)\right]}\\ &=\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}g(x+h)}+\lim_{h\to 0}{\frac{g(x+h)-g(x)}{h}f(x)}\\ &=f'(x)g(x)+f(x)g'(x). \end{aligned}\]Therefore $D(fg)(x)=k’(x)=f’(x)g(x)+f(x)g’(x)$, and $D(fg)=(Df)g+f(Dg)$. $\square$
Exercise 2.2.3. Prove that $D(\sin)=\cos$.
Proof. Let $f=\sin$. Then
We have $\sin (x+h)=\sin x \cos h + \cos x \sin h$. Thus
\[\begin{aligned} f'(x) &=\lim_{h\to 0}{\frac{\sin (x+h)-\sin x}{h}}\\ &=\lim_{h\to 0}{\frac{\sin x (\cos h - 1)}{h}} + \lim_{h\to 0}{\frac{\cos x\sin h}{h}}\\ &=\sin x \lim_{h\to 0}{\frac{ \cos h - 1}{h}} + \cos x \lim_{h\to 0}{\frac{\sin h}{h}}\\ &=\cos x. \end{aligned}\]Therefore $f’=\cos$, and $D(\sin)=\cos$. $\square$