Derivatives

Donovan D Chang

2 Derivatives

The definition of a derivative can be given in two ways. Simply as the slope of a tangent line, or as the instantaneous rate of change in the dependent variable to the change of the independent variable. By saying instantaneous, we take the limit so that the interval of the independent variable be infinitesimal.

Formal definitions [10] are as follows:

Definition. (Derivative)
The derivative of a function $f$ at a number $a$ , denoted $f'(a)$ , is defined as

(1) $\begin{equation*} f'(a)=\lim_{h\to 0} \frac{f(a+h)-f(a)}{h} \end{equation*}$

given the limit is defined.

Alternatively, the derivative can also be defined as the slope $m$ of the tangent line to the curve $y=f(x)$ at the point $P(a,f(a))$ :

(2) $\begin{equation*} m=f'(a)=\lim_{x\to a} \frac{f(x)-f(a)}{x-a} \end{equation*}$

again, given the limit is defined.

Note that the definitions (1) and (2) are equivalent. By letting $h=x-a$ , the equation (2) is obtained from the equation (1). In both cases, the derivative of $f(x)$ at $x=a$ refers to the ratio of the change in $f$ to the change in $x$ by $h$ . This is clearly shown in the following plot (Figure 4). As $h$ approaches $0$ , the blue line whose slope is the average ratio of change for the given interval $h$ near $a$ , approaches the red line whose slope is the instantaneous ratio of change of $f$ to $x$ at $x=a$ .

As was the case we expanded the notion of continuity at a point to continuity over a given interval or the whole domain of a given function, we can define the derivative of a whole function over its domain as follows [10]:

Definition. (Derivative of Function)
The derivative of a function $f(x)$ , $f'(x)$ is defined as

$f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$

given the limit is defined, i.e. exists, for $x\in \mathrm{Dom}(f)$ .

Alternative notations for expressing derivatives are introduced here.

$\begin{align*} f'(x) &=y' &&=\frac{dy}{dx} &=\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}\\ &=\frac{df}{dx} &&=\frac{d}{dx}f(x)\\ &=Df(x) &&=D_xf(x) \end{align*}$

Let us illustrate with an example.

Example. Find the derivative of $f(x)=\sqrt{x}$ .

1. Express Derivative According to Definition

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\\ \end{align*}$

However, plugging in $h=0$ would only result in $\frac{0}{0}$ , which is not defined, and thus we need to convert this into an equivalent expression by rationalization.

2. Rationalize to Convert into Form that Can Be Evaluated

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\\ &= \lim_{h \to 0} \left( \frac{\sqrt{x+h}-\sqrt{x}}{h} \right) \left(\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\\ &= \lim_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}\\ &= \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} \end{align*}$

By rationalization, the form, $\lim_{h \to 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$ , that could not have been evaluated, turned into the form that can be evaluated, $\lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}$ . Note that plugging in $h=0$ does not result in an undefined expression.

3. Complete Evaluation by Inputting $h=0$

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &= \frac{1}{\sqrt{x}+\sqrt{x}}\\ &= \frac{1}{2\sqrt{x}} \end{align*}$

Therefore, we conclude $f'(x)= \frac{1}{2\sqrt{x}}$ .

Let us have a look at another example involving a trigonometric function.

Example. Find the derivative of $f(x)=\sin x$ , given $\lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1$ .

The basic strategy is essentially the same as the example above, as the initial form by definition only leads to an undefined $\frac{0}{0}$ . However, in this example we need to utilize trigonometric identities, since we are evaluating the limit of a trigonometric function, instead of an irrational function.

1. Express Derivative According to Definition

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &= \lim_{h \to 0} \frac{\sin{(x+h)}-\sin{x}}{h}\\ \end{align*}$

Again, plugging in $h=0$ would only result in $\frac{0}{0}$ , which is not defined, and thus we need to convert this into an equivalent expression using trigonometric identities. We utilize one of sum identities, $\sin{(x+y)}=\sin x \cos y+\cos x \sin y$ .

2. Convert into Form that Can Be Evaluated with Trigonometric Identity

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{\sin{(x+h)}-\sin{x}}{h}\\ &= \lim_{h \to 0} \frac{\sin{x}\cos{h}+\cos{x}\sin{h}-\sin{x}}{h}\\ &= \lim_{h \to 0} \frac{\sin{x} \left( \cos{h}-1 \right)+\cos{x}\sin{h}}{h}\\ &= \sin{x}\lim_{h \to 0} \frac{\cos{h}-1}{h} +\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h}\\ &= \sin{x}\lim_{h \to 0} \frac{(\cos{h}-1)(\cos{h}+1)}{h(\cos{h}+1)}+\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h}\\ &= \sin{x}\lim_{h \to 0} \frac{\cos^2{h}-1}{h(\cos{h}+1)}+\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h}\\ &= \sin{x}\lim_{h \to 0} \frac{\sin^2{h}}{h(\cos{h}+1)}+\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h}\\ &= \sin{x}\lim_{h \to 0} \frac{\sin{h}}{h} \lim_{h \to 0} \frac{\sin{h}}{\cos{h}+1}+\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h} \end{align*}$

Using the sum identity of $\sin$ , the form, $\lim_{h \to 0} \frac{\sin{x+h}-\sin{x}}{h}$ , that could not have been evaluated, turned into the form that can be evaluated, $\sin{x}\lim_{h \to 0} \frac{\sin{h}}{h} \lim_{h \to 0} \frac{\sin{h}}{\cos{h}+1}+\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h}$ . Note that plugging in $h=0$ does not result in an undefined expression.

3. Use Given Information $\lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1$ .

$\begin{align*} f'(x) &= \sin{x}\lim_{h \to 0} \frac{\sin{h}}{h} \lim_{h \to 0} \frac{\sin{h}}{\cos{h}+1}+\cos{x}\lim_{h \to 0} \frac{\sin{h}}{h}\\ &= \sin{x} \lim_{h \to 0} \frac{\sin{h}}{\cos{h}+1}+\cos{x} \end{align*}$

4. Complete Evaluation by Inputting $h=0$

$\begin{align*} f'(x) &= \sin{x} \lim_{h \to 0} \frac{\sin{h}}{\cos{h}+1}+\cos{x}\\ &= \left(\sin{x}\right)\cdot \left(\frac{0}{1}\right)+\cos{x}\\ &= \cos{x} \end{align*}$

Therefore, we conclude $f'(x)= \cos x$ .

To generalize, let us introduce and explore a few rules and patterns [10] observed when differentiating diverse functions, such as polynomials, and product of multiple functions.

Theorem. (Power Rule)
For any $n \in \mathbb{R}$ ,
$\frac{d}{dx}x^n=nx^{n-1}$

Proof.
Let $f(x)=x^n$ . Then,

$\begin{align*} f'(a) &= \lim_{x \to a} \frac{f(x)-f(a)}{x-a}\\ &= \lim_{x \to a} \frac{x^n-a^n}{x-a}\\ &= \lim_{x \to a} \left(x^{n-1}+x^{n-2}a+\cdots+xa^{n-2}+a^{n-1}\right)\\ &= a^{n-1}+a^{n-2}a+\cdots+aa^{n-2}+a^{n-1}\\ &= na^{n-1} \end{align*}$

As we expanded our notion of a derivative at a point $a$ to a derivative of a function $f(x)$ , we substitute $a$ with $x$ . Therefore,

$f'(x)=\frac{d}{dx}x^n=nx^{n-1}$

Theorem. (Sum Rule)
If $f$ and $g$ are both differentiable, then
$\frac{d}{dx}\left[f(x)+g(x)\right]=\frac{d}{dx}f(x)+\frac{d}{dx}g(x)=f'(x)+g'(x)$

Proof.
Let $F(x)=f(x)+g(x)$ . Then,

$\begin{align*} F'(x) &= \lim_{h \to 0} \frac{F(x+h)-F(x)}{h}\\ &= \lim_{h \to 0} \left[\frac{\left[f(x+h)+g(x+h)\right]-\left[f(x)-g(x)\right]}{h}\right]\\ &= \lim_{h \to 0} \left[\frac{f(x+h)-f(x)}{h}+\frac{g(x+h)-g(x)}{h}\right]\\ &= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}+\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}\\ &= f'(x)+g'(x) \end{align*}$

Theorem. (Product Rule)
If $f$ and $g$ are both differentiable, then
$\frac{d}{dx}\left[f(x)g(x)\right]=f(x)\frac{d}{dx}g(x)+g(x)\frac{d}{dx}f(x)=f(x)g'(x)+g(x)f'(x)$

Proof.
Let $F(x)=f(x)g(x)$ . Then,

$\begin{align*} F'(x) &= \lim_{h \to 0} \frac{F(x+h)-F(x)}{h}\\ &= \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}\\ &= \lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h}\\ &= \lim_{h \to 0} \frac{f(x+h)\left[g(x+h)-g(x)\right]+g(x)\left[f(x+h)-f(x)\right]}{h}\\ &= \lim_{h \to 0} \frac{f(x+h)\left[g(x+h)-g(x)\right]}{h}+\lim_{h \to 0}\frac{g(x)\left[f(x+h)-f(x)\right]}{h}\\ &= f(x)g'(x)+f'(x)g(x) \end{align*}$

Chain Rule

The chain rule concerns the case of a composite function.

Theorem. (Chain Rule)
If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite function $F=\left(f\circ g\right)(x)=f(g(x))$ is differentiable at $x$ , and
$F'(x)=f'(g(x))\cdot g'(x)$

An alternative expression can be obtained by letting $y=f(u)$ and $u=g(x)$
$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

Let us illustrate with an example.

Example. Find $F'(x)$ , where $F(x)=\sqrt{x^2-1}$ .

Let $f(u)=\sqrt{u}$ and $u=g(x)=x^2-1$ . Then,

$\frac{d}{du}f(u) =\frac{1}{2}u^{-\frac{1}{2}} =\frac{1}{2\sqrt{u}} \;\; \text{and} \;\; \frac{du}{dx} =\frac{d}{dx}g(x) =2x$

Therefore,

$\begin{align*} F'(x) &= \frac{d}{du}f(u)\frac{du}{dx}\\ &= \frac{1}{2\sqrt{u}}\cdot 2x\\ &= \frac{x}{\sqrt{x^2-1}} \end{align*}$

Here is another example involving an exponential function.

Example. Find $F'(x)$ , where $F(x)=e^{x^2}$ .

Let $f(u)=e^u$ and $u=g(x)=x^2$ . Then,

$\frac{d}{du}f(u)=e^u \;\; \text{and} \;\; \frac{du}{dx} =\frac{d}{dx}g(x)=2x$

Therefore,

$\begin{align*} F'(x) &= \frac{d}{du}f(u)\frac{du}{dx}\\ &= e^{u} \cdot 2x\\ &= 2xe^{x^2} \end{align*}$

Implicit differentiation

So far we have only dealt with the differentiation of functions in the form of $y=f(x)$ . However, at times, it is difficult or cumbersome to express the function in the form of $y=f(x)$ , but instead, defined implicitly in the for of $f(x,y)=0$ , as opposed to explicitly. Circle, for example, $x^2+y^2=r^2$ , is a notable example.

Example. Find $\frac{dy}{dx}$ , where $x^2+y^2=r^2$ .

$\begin{align*} x^2+y^2 &= r^2\\ \frac{d}{dx}(x^2+y^2) &= \frac{d}{dx}(r^2)\\ \frac{d}{dx}(x^2)+\frac{d}{dx}(y^2) &= 0\\ 2x+\frac{dy}{dx}\frac{d}{dy}(y^2) &= 0\\ 2x+2y\frac{dy}{dx} &= 0 \end{align*}$

Therefore,

$\frac{dy}{dx}=-\frac{x}{y}$

Another way of approaching the example above would be finding the explicit expression of $y$ in terms of $x$ , and then differentiate. However, it would have been much more complex and prone to error involving differentiation of a composite square root function. Therefore, implicit differentiation can be a useful technique to avoid expensive computations.

2 Derivatives

Chain Rule

Implicit differentiation

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