Differentiation

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The essence of ``differentiation'' is approximating arbitrary functions by linear functions.



Definition (Differentiability and derivative)

Let \(f(x)\) be a function defined around \(x=a\). We say \(f(x)\) is differentiable at \(x=a\) if the following limit exists:

\[\lim_{x\to a}\frac{f(x) - f(a)}{x - a} = \lim_{h\to 0}\frac{f(a+h) - f(a)}{h}.\]

This limit value is called the differential coefficient of \(f(x)\) at \(x=a\), and denoted by

\[f'(a) \text{ or } \frac{df}{dx}(a) \text{ or } \frac{d}{dx}f(a).\]

If \(f(x)\) is defined on an open interval \(I\) and is differentiable at all points in \(I\), then \(f(x)\) is said to be differentiable on \(I\). In this case, we can make correspondence between each \(c\in I\) and \(f'(c)\), which defines a function on \(I\). Such a function is called the derivative of \(f(x)\) and denoted by

\[f'(x) \text{ or } \frac{df}{dx}(x) \text{ or } \frac{d}{dx}f(x).\]

Remark. We also use the verb ``differentiate'' to mean the act of finding the derivative of a function. □

Example. Let \(f(x) = x^2\) defined on \(\mathbb{R}\). Its derivative is given by

\[ \begin{eqnarray} \frac{(x + h)^2 - x^2}{h} &=& \frac{2hx + h^2}{h}\\ &=& 2x + h\\ &\to& 2x ~ (h \to 0). \end{eqnarray} \]

Therefore, \[\frac{dx^2}{dx} = 2x.\]

□

Example. Let us find the derivative of \(\sin x\). Thus we need to find the following limit

\[\lim_{h\to 0}\frac{\sin(x + h) - \sin x}{h}.\]

Note the following formula

\[\sin\alpha -\sin\beta = 2\cos\frac{\alpha + \beta}{2}\sin\frac{\alpha - \beta}{2}.\]

(You should prove this.) By setting \(\alpha = x + h\) and \(\beta = x\), we have

\[ \begin{eqnarray} \frac{\sin(x + h) - \sin x}{h} &=& 2\cos\left(x + \frac{h}{2}\right)\frac{\sin\frac{h}{2}}{h}\\ &\to& \cos x\cdot 1 ~ (h \to 0) \end{eqnarray} \]

where we have used the fact \(\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1\) with \(\theta = \frac{h}{2}\).

Similarly, we can derive

\[\frac{d\cos x}{dx} = -\sin x.\]

(Exercise!) □

Example. Let's prove 

\[\frac{d}{dx}e^x = e^x.\]

In fact, recalling that

\[\lim_{h\to 0}\frac{e^h - 1}{h} = 1,\]

we have

\[\frac{e^{x + h} - e^x}{h} = e^x\frac{e^h - 1}{h} \to e^x ~ (h \to 0).\]

□

See also: Elementary functions for the limit \((e^h - 1)/h \to 1\). □

If \(f(x)\) is differentiable at \(x=a\), \(f'(a)\) corresponds to the slope of the tangent line of \(y=f(x)\) at \((a, f(a))\). The equation of the tangent line at \((a, f(a))\) is given by

\[y - f(a) = f'(a)(x - a).\]

Theorem (Differentiable functions are continuous)

If the function \(f(x)\) is differentiable at \(x=a\), then it is continuous at \(x=a\).

Proof. Suppose \(f(x)\) is differentiable at \(x=a\). Then \(f'(a)\) exists. Therefore

\[ \begin{eqnarray} \lim_{x\to a}f(x) &=& \lim_{x\to a}\left[\frac{f(x) - f(a)}{(x-a)}\cdot (x-a) + f(a)\right]\\ &=& f'(a)\cdot 0 + f(a) = f(a). \end{eqnarray} \]

Hence \(f(x)\) is continuous at \(x=a\). ■

Remark. The converse of this theorem is not necessarily true. See the following example. □

Example. Consider \(f(x) = |x|\).

\[\lim_{x\to +0}|x| = \lim_{x\to -0}|x| = f(0) = 0\]

so \(f(x)\) is continuous at \(x = 0\). However,

\[\lim_{x\to +0}\frac{f(x) - f(0)}{x - 0} = \lim_{x\to +0}\frac{x - 0}{x - 0} = 1\]

whereas

\[\lim_{x\to -0}\frac{f(x) - f(0)}{x - 0} = \lim_{x\to -0}\frac{-x - 0}{x - 0} = -1.\]

Thus, the limit \(\lim_{x\to 0}\frac{f(x) - f(0)}{x - 0}\) does not exist. This means that \(f(x) = |x|\) is not differentiable at \(x = 0\). □



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