Properties of differentiation

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 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) \[\frac{d}{dx}[kf(x) + lg(x)] = kf'(x) + lg'(x)\] where \(k, l\) are constants. 
  2. (Product rule) \[\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x).\] 
  3. (Quotient rule) \(\frac{f(x)}{g(x)}\) is differentiable on \(\{x \mid g(x) \neq 0, x\in I\}\) and \[\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.\]
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 \(a\in I\), \[\begin{eqnarray*} \frac{f(x)g(x) - f(a)g(a)}{x - a} &=& \frac{f(x)g(x) - f(x)g(a) + f(x)g(a) - f(a)g(a)}{x - a}\\ &=& f(x)\cdot\frac{g(x) - g(a)}{x-a} + \frac{f(x) - f(a)}{x-a}\cdot g(a)\\ &\to& f(a)g'(a) + f'(a)g(a) ~ (x \to a). \end{eqnarray*}\]
  3. Note that \(f(x)\) and \(g(x)\) are both continuous at \(x = a\). For any \(a\in I\), \[ \begin{eqnarray*} \frac{f(x)}{g(x)} - \frac{f(a)}{g(a)} &=& \frac{f(x)g(a) - f(a)g(x)}{g(x)g(a)}\\ &=& \frac{[f(x) - f(a)]g(x) - f(x)[g(x) - g(a)]}{g(x)g(a)}. \end{eqnarray*}\] The rest is exercise.
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Example. \(\tan x\) is differentiable for \(x \neq \left(n + \frac{1}{2}\right)\pi\) and
\[\frac{d}{dx}\tan x = \frac{1}{\cos^2 x}.\]
In fact,
\[\begin{eqnarray*} \frac{d}{dx}\tan x &=& \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)\\ &=& \frac{(\sin x)'\cos x - \sin x(\cos x)'}{\cos^2 x}\\ &=& \frac{\cos^2x + \sin^2x}{\cos^2 x} = \frac{1}{\cos^2x}. \end{eqnarray*}\]
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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
\[(g\circ f)'(x) = g'(f(x))f'(x).\]
Proof. For any \(a\in I\), we have
\[\begin{eqnarray} \frac{(g\circ f)(x) - (g\circ f)(a)}{x - a} &=& \frac{g(f(x)) - g(f(a))}{x - a}\\ &=&\frac{g(f(x)) - g(f(a))}{f(x) - f(a)}\cdot\frac{f(x) - f(a)}{x - a}. \end{eqnarray} \]
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 \(\frac{g(y) - g(b)}{y - b} \to g'(b) = g'(f(a))\) as \(y \to b\). Thus
\[\lim_{x\to a}\frac{g(f(x)) - g(f(a))}{f(x) - f(a)} = g'(f(a)).\]
\(f(x)\) is differentiable at \(x=a\) and
\[\lim_{x\to a}\frac{f(x) - f(a)}{x - a} = f'(a).\]
Therefore,
\[\lim_{x\to a}\frac{(g\circ f)(x) - (g\circ f)(a)}{x-a} = g'(f(a))f'(a).\]
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
\[\frac{d}{dx}f(ax) = \frac{d}{dy}f(y)\cdot\frac{dy}{dx} = f'(y)\cdot a = af'(ax).\]
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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
\[\frac{d}{dy}f^{-1}(y) = \frac{1}{f'(f^{-1}(y))}.\]
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
\[\frac{d}{dy}f^{-1}(y) = \frac{1}{f'(x)} = \frac{1}{f'(f^{-1}(y))}.\]
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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 \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). Then \(x = \arcsin y\). From the above theorem,
\[\frac{d}{dy}\arcsin y = \frac{1}{(\sin x)'} = \frac{1}{\cos x}.\]
Since \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), \(\cos x > 0\) 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
\[\begin{eqnarray} \frac{d}{dx}x^a &=& \frac{d}{dy}e^y\cdot\frac{d}{dx}(a\log x) \\ &=& e^y\cdot\frac{a}{x}\\ &=& e^{a\log x}\cdot\frac{a}{x}\\ &=& x^a\cdot\frac{a}{x}\\ &=& ax^{a-1}. \end{eqnarray} \]
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Let \(f(x)\) be a differentiable function on an open interval \(I\). Then its derivative \(f'(x)\) is also a function on \(I\). We may consider the differentiability of \(f'(x)\), too.

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

Example (eg:x2). Consider the function \(f(x)\) defined by
\[ f(x) = \left\{ \begin{array}{cc} x^2 & (x \geq 0),\\ -x^2 & (x < 0). \end{array}\right. \]
We can see that \(f'(x) = 2x\) for \(x > 0\) and \(f'(x) = -2x\) for \(x < 0\). Is \(f'(x)\) 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,
\[\lim_{x\to 0}\frac{f(x) - f(0)}{x - 0} = 0\]
so that \(f(x)\) is differentiable at \(x =0\) and \(f'(0) = 0\). We can summarize the derivative as
\[f'(x) = 2|x|\]
for all \(x \in \mathbb{R}\).

However, \(f'(x) = 2|x|\) 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 \(f'(x)\) is also differentiable, its derivative \((f'(x))'\) is called the second order derivative or simply second derivative and is denoted
\[f''(x) \text{ or } \frac{d^2f}{dx^2}(x) \text{ or } \frac{d^2}{dx^2}f(x).\]
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
\[f^{(n)}(x) \text{ or } \frac{d^nf}{dx^n}(x) \text{ or } \frac{d^n}{dx^n}f(x).\]
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 \(f', f'', \cdots, f^{(n)}\) exist and are continuous.
  2. The function \(f(x)\) is said to be infinitely differentiable or smooth or of class \(C^{\infty}\) 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 \(C^{\infty}\). □

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



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