Properties of differentiation

 We show some basic properties of differentiation, such as linearity, the product rule, the quotient rule, the chain rule, etc. We also introduce higher-order derivatives and differentiability classes.

Theorem (Properties of differentiation)

Let $f(x)$ and $g(x)$ be differentiable functions on an open interval $I$.

1. (Linearity) $$\frac{d}{dx}[kf(x)+lg(x)]=k{f}^{\prime }(x)+l{g}^{\prime }(x)$$ where $k,l$ are constants. 
2. (Product rule) $$\frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x).$$ 
3. (Quotient rule) $\frac{f(x)}{g(x)}$ is differentiable on $\{x\mid g(x)\ne 0,x\in I\}$ and $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{[g(x){]}^2}.$$

Proof. 

1. Exercise.
2. Since $f(x)$ is differentiable at $x=a$, it is continuous at $x=a$ so $f(x)\to f(a)$ as $x\to a$. For any $a\in I$, $$\begin{array}{rcl}\frac{f(x)g(x)-f(a)g(a)}{x-a}& =& \frac{f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)}{x-a}\\ & =& f(x)\cdot \frac{g(x)-g(a)}{x-a}+\frac{f(x)-f(a)}{x-a}\cdot g(a)\\ & \to & f(a)g^{\prime }(a)+f^{\prime }(a)g(a)(x\to a).\end{array}$$
3. Note that $f(x)$ and $g(x)$ are both continuous at $x=a$. For any $a\in I$, $$\begin{array}{rcl}\frac{f(x)}{g(x)}-\frac{f(a)}{g(a)}& =& \frac{f(x)g(a)-f(a)g(x)}{g(x)g(a)}\\ & =& \frac{[f(x)-f(a)]g(x)-f(x)[g(x)-g(a)]}{g(x)g(a)}.\end{array}$$ The rest is exercise.

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Example. $\tan x$ is differentiable for $x\ne (n+\frac{1}{2})\pi$ and

$$\frac{d}{dx}\tan x=\frac{1}{{\cos }^{2}x}.$$

In fact,

$$\begin{array}{rcl}\frac{d}{dx}\tan x& =& \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)\\ & =& \frac{(\sin x{)}^{\prime }\cos x-\sin x(\cos x{)}^{\prime }}{{\cos }^{2}x}\\ & =& \frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}.\end{array}$$

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Theorem (Chain rule)

Let $f(x)$ be a differentiable function on an open interval $I$ and $g(x)$ be a differentiable function on an open interval $J$. Suppose that for all $x\in I$, $f(x)\in J$. Then the function composition $(g\circ f)(x)$ is differentiable on $I$ and its derivative is given by

$$(g\circ f{)}^{\prime }(x)=g^{\prime }(f(x))f^{\prime }(x).$$

Proof. For any $a\in I$, we have

$$\begin{array}{rcl}\frac{(g\circ f)(x)-(g\circ f)(a)}{x-a}& =& \frac{g(f(x))-g(f(a))}{x-a}\\ & =& \frac{g(f(x))-g(f(a))}{f(x)-f(a)}\cdot \frac{f(x)-f(a)}{x-a}.\end{array}$$

Since $f(x)$ is differentiable at $x=a$, it is continuous at $x=a$. Hence $f(x)\to f(a)$ as $x\to a$.

Let $b=f(a)\in J$. $g(y)$ is differentiable at $y=b$ so that $\frac{g(y)-g(b)}{y-b}\to g^{\prime }(b)=g^{\prime }(f(a))$ as $y\to b$. Thus

$$\underset{x\to a}{lim}\frac{g(f(x))-g(f(a))}{f(x)-f(a)}=g^{\prime }(f(a)).$$

$f(x)$ is differentiable at $x=a$ and

$$\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=f^{\prime }(a).$$

Therefore,

$$\underset{x\to a}{lim}\frac{(g\circ f)(x)-(g\circ f)(a)}{x-a}=g^{\prime }(f(a))f^{\prime }(a).$$

In particular, $f(x)$ is differentiable at $x=a$. ■

Remark. If we set $y=f(x)$ and $z=g(y)$, we can express the chain rule as 

$$\frac{dz}{dx}=\frac{dz}{dy}\cdot \frac{dy}{dx}.$$

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Example. Let $f(x)$ be a differentiable function and $a$ be an arbitrary constant. The function $f(ax)$ may be construed as a composition of $y=ax$ and $z=f(y)$. Therefore, its derivative is obtained as

$$\frac{d}{dx}f(ax)=\frac{d}{dy}f(y)\cdot \frac{dy}{dx}=f^{\prime }(y)\cdot a=a{f}^{\prime }(ax).$$

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Theorem (Derivative of inverse)

Let $y=f(x)$ be a function on an open interval $I$ that is differentiable and has the inverse $f^{-1}(y)$. $f^{-1}(y)$ is differentiable at $y=f(x)$ and the derivative is given by

$$\frac{d}{dy}{f}^{-1}(y)=\frac{1}{f^{\prime }(f^{-1}(y))}.$$

Proof. Let $y=f(x)$. Then,

$$x=f^{-1}(y)=f^{-1}(f(x))=(f^{-1}\circ f)(x).$$

Differentiating both sides with respect to $x$ and applying the chain rule to the right-hand side, we have

$$1=\frac{d}{dy}{f}^{-1}(y)\cdot \frac{d}{dx}f(x).$$

From this, we conclude

$$\frac{d}{dy}{f}^{-1}(y)=\frac{1}{f^{\prime }(x)}=\frac{1}{f^{\prime }(f^{-1}(y))}.$$

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Example. Let us show

$$\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}(-1<x<1).$$

Let $y=\sin x$ for $x\in (-\frac{\pi }{2},\frac{\pi }{2})$. Then $x=\arcsin y$. From the above theorem,

$$\frac{d}{dy}\arcsin y=\frac{1}{(\sin x{)}^{\prime }}=\frac{1}{\cos x}.$$

Since $x\in (-\frac{\pi }{2},\frac{\pi }{2})$, $\cos x>0$ so that we can write

$$\cos x=\sqrt{1-{\sin }^{2}x}.$$

By substituting $y^2={\sin }^{2}x$, we have

$$\frac{d}{dy}\arcsin y=\frac{1}{\sqrt{1-y^2}}.$$

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Example. Consider $f(x)=x^a$ where $a\in \mathbb{R}$ is arbitrary and $x>0$. Since

$$x^a=e^{a\log x},$$

we may consider $x^a$ as a composition of $y=a\log x$ and $z=e^y$. Therefore its derivative is

$$\begin{array}{rcl}\frac{d}{dx}{x}^a& =& \frac{d}{dy}{e}^y\cdot \frac{d}{dx}(a\log x)\\ & =& e^y\cdot \frac{a}{x}\\ & =& e^{a\log x}\cdot \frac{a}{x}\\ & =& x^a\cdot \frac{a}{x}\\ & =& a{x}^{a-1}.\end{array}$$

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Let $f(x)$ be a differentiable function on an open interval $I$. Then its derivative $f^{\prime }(x)$ is also a function on $I$. We may consider the differentiability of $f^{\prime }(x)$, too.

Example. If $f(x)$ is a polynomial function, then it is differentiable on the entire $\mathbb{R}$ and its derivative $f^{\prime }(x)$ is also a polynomial function. Therefore $f^{\prime }(x)$ is also differentiable on the entire $\mathbb{R}$. □

Example (eg:x2). Consider the function $f(x)$ defined by

$$f(x)=\{\begin{array}{cc}{x}^2& (x\ge 0),\\ -x^2& (x<0).\end{array}$$

We can see that $f^{\prime }(x)=2x$ for $x>0$ and $f^{\prime }(x)=-2x$ for $x<0$. Is $f^{\prime }(x)$ defined at $x=0$?

If $x>0$, then

$$\frac{f(x)-f(0)}{x-0}=\frac{x^2-0}{x}=x\to 0(x\to +0).$$

If $x<0$, then

$$\frac{f(x)-f(0)}{x-0}=\frac{-x^2-0}{x}=-x\to 0(x\to -0).$$

Therefore,

$$\underset{x\to 0}{lim}\frac{f(x)-f(0)}{x-0}=0$$

so that $f(x)$ is differentiable at $x=0$ and $f^{\prime }(0)=0$. We can summarize the derivative as

$$f^{\prime }(x)=2|x|$$

for all $x\in \mathbb{R}$.

However, $f^{\prime }(x)=2|x|$ is not differentiable at $x=0$ as we have seen before. □

Definition (Higher order derivatives)

Let $f(x)$ be a differentiable function. If the derivative $f^{\prime }(x)$ is also differentiable, its derivative $(f^{\prime }(x){)}^{\prime }$ is called the second order derivative or simply second derivative and is denoted

$$f^{″}(x)\text{ or }\frac{d^{2}f}{d{x}^2}(x)\text{ or }\frac{d^2}{d{x}^2}f(x).$$

Similarly, we can define the third derivative, forth derivative, and so on. In general, the derivative obtained by differentiating $f(x)$ $n$ times is called the $n$-th (order) derivative of $f(x)$ and is denoted

$$f^{(n)}(x)\text{ or }\frac{d^{n}f}{d{x}^n}(x)\text{ or }\frac{d^n}{d{x}^n}f(x).$$

Remark. It is often convenient to consider the function $f(x)$ itself to be the ``$0$-th derivative'' and write $f^{(0)}(x)$. □

Definition (Differentiability classes)

Let $f(x)$ be a function on an open interval $I$. Let $n$ be a non-negative integer ($n=0,1,2,\cdots$).

1. The function $f(x)$ is said to be of class $C^n$ if the derivatives $f^{\prime },f^{″},\cdots ,f^{(n)}$ exist and are continuous.
2. The function $f(x)$ is said to be infinitely differentiable or smooth or of class $C^{\mathrm{\infty }}$ if it has derivatives of all orders. 

Example. $f(x)=|x|$ is continuous on all $\mathbb{R}$, but not differentiable at $x=0$. Therefore it is of class $C^0$. □

Example. Polynomial functions, $\sin x$, $\cos x$, $e^x$ are of $C^{\mathrm{\infty }}$. □

Example. The function in the above example (eg:x2) is of $C^1$, but not of $C^2$. □