Differentiation

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:

$$\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}=\underset{h\to 0}{lim}\frac{f(a+h)-f(a)}{h}.$$

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

$$f^{\prime }(a)\text{ or }\frac{df}{dx}(a)\text{ or }\frac{d}{dx}f(a).$$

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 $f^{\prime }(c)$, which defines a function on $I$. Such a function is called the derivative of $f(x)$ and denoted by

$$f^{\prime }(x)\text{ or }\frac{df}{dx}(x)\text{ or }\frac{d}{dx}f(x).$$

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

$$\begin{array}{rcl}\frac{(x+h{)}^2-x^2}{h}& =& \frac{2hx+h^2}{h}\\ & =& 2x+h\\ & \to & 2x(h\to 0).\end{array}$$

Therefore, $$\frac{d{x}^2}{dx}=2x.$$

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

$$\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}.$$

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

$$\begin{array}{rcl}\frac{\sin (x+h)-\sin x}{h}& =& 2\cos (x+\frac{h}{2})\frac{\sin \frac{h}{2}}{h}\\ & \to & \cos x\cdot 1(h\to 0)\end{array}$$

where we have used the fact $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1$ 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

$$\underset{h\to 0}{lim}\frac{e^h-1}{h}=1,$$

we have

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

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

If $f(x)$ is differentiable at $x=a$, $f^{\prime }(a)$ 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

$$y-f(a)=f^{\prime }(a)(x-a).$$

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 $f^{\prime }(a)$ exists. Therefore

$$\begin{array}{rcl}\underset{x\to a}{lim}f(x)& =& \underset{x\to a}{lim}[\frac{f(x)-f(a)}{(x-a)}\cdot (x-a)+f(a)]\\ & =& f^{\prime }(a)\cdot 0+f(a)=f(a).\end{array}$$

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

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

Example. Consider $f(x)=|x|$.

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

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

$$\underset{x\to +0}{lim}\frac{f(x)-f(0)}{x-0}=\underset{x\to +0}{lim}\frac{x-0}{x-0}=1$$

whereas

$$\underset{x\to -0}{lim}\frac{f(x)-f(0)}{x-0}=\underset{x\to -0}{lim}\frac{-x-0}{x-0}=-1.$$

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