## Increments

Recall that the fact that $$y$$ is a function of $$x$$ is expressed by the equation $$y=f(x)$$, and the particular value of the function when $$x$$ has a definite value $$x_{0}$$ is then expressed as $$f(x_{0})$$.

When $$y=f(x)$$, it is, in general, true that a change in $$x$$ causes a change in the dependent variable $$y$$, and that if the change in $$x$$ is sufficiently small, the change in $$y$$ is small also. Some exceptions to this may be noticed later, but this is the general rule. A change in $$x$$ is called an increment of $$x$$ and is denoted by the symbol $$\Delta x$$ (read delta x. You are warned against reading this symbol delta times x, it having no such meaning). Similarly, a change in $$y$$ is called an increment of $$y$$ and is denoted by $$\Delta y$$. For example, consider $y=x^{2}+3x+2.$ When $$x=2$$, $$y=12$$. When $$x=2.1$$, $$y=12.71$$. The change in $$x$$ is 0.1, and the change in $$y$$ is 0.71, and we write $\Delta x=0.1,\quad\Delta y=0.71.$

So, in general, if $$x_{0}$$ is one value of $$x$$, and $$x_{1}$$ a second value of $$x$$, then $\Delta x=x_{1}-x_{0},\quad or\quad x_{1}=x_{0}+\Delta x;$ and if $$y_{0}$$ and $$y_{1}$$ are the corresponding values of $$y$$ (i.e. $$y_{0}=f(x_{0})$$ and $$y_{1}=f(x_{1})$$), then $\Delta y=y_{1}-y_{0},\quad or\quad y_{1}=y_{0}+\Delta y.$

The word increment really means “increase”, but as we are dealing with algebraic quantities, the increment may be negative when it means a decrease. For example, if you invest \$1000 and at the end of a year have$1200, the increment of your wealth is \$200. If you have \$800 at the end of the year, the increment is –\200. So, if a thermometer registers $$32^{\circ}$$ in the morning and $$25^{\circ}$$ at night, the increment is $$-7^{\circ}$$. • The increment is always the second value of the quantity considered minus the first value. Example 1 If $$y=\sqrt{x}$$, find the increment $$\Delta y$$ at $$x=0.5$$ for $$\Delta x=0.01$$. Solution Let $$x_{0}=0.5$$. Then $$x_{1}=x_{0}+\Delta x=0.51$$. The corresponding values of $$y$$ are \begin{aligned} y_{0} & =\sqrt{x_{0}}=\sqrt{0.5}\approx0.7071\\ y_{1} & =\sqrt{x_{1}}=\sqrt{0.51}\approx0.7141\end{aligned} Thus $\Delta y=y_{1}-y_{0}\approx0.007.$ Now, having determined increments of $$x$$ and of $$y$$, the next step is to compare them by dividing the increment of $$y$$ by the increment of $$x$$. This is what we do when finding average velocity; we divide an increment of distance by an increment of time. In finding average acceleration, we divide an increment of velocity by an increment of time. ## Average Velocity Versus Instantaneous (or True) Velocity #### Read about the difference between average velocity and instantaneous velocity If a car travels a distance of 240 kilometers in 3 hours, we say it has traveled at the rate of 80 kilometers an hour (or 80 kilometers per hour). But we know that this does not necessarily mean that the speedometer registers 80 km/hr all the time. 80 km/hr is its average speed (or average velocity). In general, suppose that an object is moving in a straight line. Let’s choose one direction as positive and the opposite as negative, and one point as the origin $$O$$. Let $$s$$ be the object’s position (i.e., its coordinate) on this straight line, and $$s=f(t)$$ be a function denoting the position of the object at time $$t$$. The object’s average velocity during an interval of time $$[t_{0},t_{1}]$$ is found by dividing $$\Delta s=f(t_{1})-f(t_{0})$$ by $$\Delta t=t_{1}-t_{0}$$: \begin{aligned} v_{\text{avg}} & =\frac{f(t_{1})-f(t_{0})}{t_{1}-t_{0}}\\ & =\frac{\Delta s}{\Delta t}.\end{aligned} Because $$t_{1}=t_{0}+\Delta t$$, we can rewrite the above equation as $v_{\text{avg}}=\frac{f(t_{0}+\Delta t)-f(t_{0})}{\Delta t}.$ • Velocity vs. Speed: Velocity and speed have two distinct meanings. The average velocity is calculated by dividing change in position (called displacement)\Delta s$by change in time$\Delta t$v_{\text{avg}}=\frac{\text {displacement}}{\text{travel time}}=\frac{\text{final position}-\text {initial position}}{\text {travel time}}$ Velocity is a quantity with a sign, meaning it can be positive or negative. $\text{finial position < initial position} \Rightarrow v_{\text{avg}}<0.$ The average speed is calculated by dividing the distance that object travels by the elapsed time: $\text{average speed}=\frac{\text {distance travel}}{\text{travel time}}.$Speed is always nonnegative. If you travel to a city that is 150 kilometers away and are back in 4 hours, your average speed is (150+150)/4 = 75 km/hr, but your average velocity is zero because your final position is the same as your initial position. Example 2 Assume you drop a stone from rest and the air resistance is negligible. If $$s$$ denotes stone’s position fallen after $$t$$ seconds, then $s=\frac{1}{2}gt^{2},$ where $$g$$ is a constant called gravitational acceleration. If $$s$$ is measured in feet and $$t$$ in seconds, $$g\approx16$$. Find the average velocity of the stone (a) during the first 3 seconds of fall; (b) during the one second interval between second two and second three. Solution (a) The positin of the stone 3 seconds after the realse is $s(3)=16(3)^{2}=144$ The average velocity during the first 3 seconds is \begin{aligned} v_{\text{avg}} & =\frac{s(3)-s(0)}{3-0}\\ & =\frac{144-0}{3-0}\\ & =48\ \text{ft/s}\end{aligned} (b) The average velocity from second 2 to second 3 is \begin{aligned} v_{\text{avg}} & =\frac{s(3)-s(2)}{3-2}\\ & =\frac{16(3)^{2}-16(2)^{2}}{3-2}\\ & =16(9-4)\\ & =80\ \text{ft/s}.\end{aligned} But what is the velocity at a single instant $$t_{0}$$, instead of the average velocity over a time interval? This velocity is a physical quantity that can be measured (for example by a speedometer for a car) and because it refers to a single instant is called “instantaneous velocity.” Our intuition suggests that the instantaneous velocity is approximately equal to the average velocity if the “averaging time” $$\Delta t$$ is small. This approximation gets better and better as $$\Delta t$$ gets smaller and smaller. So for an object with the position function $$s=f(t)$$, the instantaneous velocity at $$t_{0}$$ is $v(t_{0})=\lim_{\Delta t\to0}v_{\text{avg}}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$ or $v(t_{0})=\lim_{\Delta t\to0}\frac{f(t_{0}+\Delta t_{0})-f(t_{0})}{\Delta t}.$ The above limit is called the derivative of the position function with respect to time. Example 3 As in previous example, the stone’s position at time $$t$$ is given by $s=f(t)=16t^{2}.$ What is the (instantaenous) velocity of the stone at $$t=3$$ sec? Solution First, let’s estimate it numerically and then calculate it by the methods we learned in the previous chapter. It seems that the average velocity approaches 96 ft/s as $$\Delta t\to0$$ and thus the instantaneous velocity of the stone at $$t_{0}=3$$ sec is 96 ft/s. Now let’s confirm it algebraically: \begin{aligned} \frac{\Delta s}{\Delta t} & =\lim_{\Delta t\to0}\frac{f(3+\Delta t)-f(3)}{\Delta t}\\ & =\lim_{\Delta t\to0}\frac{16(3+\Delta t)^{2}-16(3)^{2}}{\Delta t}\\ & =\lim_{\Delta t\to0}\frac{16(3^{2}+2\cdot3\cdot\Delta t+(\Delta t)^{2})-16(3)^{2}}{\Delta t}\\ & =\lim_{\Delta t\to0}\frac{16\cdot2\cdot3\cdot\Delta t+16(\Delta t)^{2}}{\Delta t}\\ & =\lim_{\Delta t\to0}[16\cdot2\cdot3+16\cdot\Delta t]\\ & =16\cdot2\cdot3+16\cdot0\\ & =96.\end{aligned} ## Increment Quotient and the Derivative If $$y=f(x)$$, the average change of $$y=f(x)$$ with respect to $$x$$ over the interval $$[x_{0},x_{1}]$$ (or $$[x_{1},x_{0}]$$ if $$x_{1}<x_{0}$$) is $\frac{\Delta y}{\Delta x}=\frac{\text{increment of }y}{\text{increment of }x}=\frac{\text{change in }y}{\text{change in }x},$ where $$\Delta y=f(x_{1})-f(x_{0})$$ and $$\Delta x=x_{1}-x_{0}$$. The limit of $$\Delta y/\Delta x$$ as $$\Delta x$$ approaches zero is called the derivative of y with respect to x at x0, and is denoted by $${\displaystyle \frac{dy}{dx}}(x_0)$$. We have then $\frac{dy}{dx}(x_0)=\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\to0}\frac{\text{change in }y}{\text{change in }x}.$ • At present, we take the symbol $$dy/dx$$ not as a fraction but as one undivided symbol to represent the derivative. Later, we will consider what meaning may be given to $$dx$$ and $$dy$$, separately. At this stage, the form $$dy/dx$$ suggests simply the fraction $$\Delta y/\Delta x$$, which has approached a definite limiting value. $$dy/dx$$ is read “dee why by dee eks” or “dee why over dee eks” and called the Leibniz notation for derivatives. Definition 1: A functiony=f(x)$is said to have a derivative for$x=x_{0}$(or to be differentiable at$x=x_{0}$), if the limit $\lim_{\Delta x\to0}\frac{f(x_{0}+\Delta x)-f(x_{0})}{\Delta x}$ exists. This limit is denoted by$f'(x_{0})$or$\dfrac{dy}{dx}(x_0)$. If$f(x)$has a derivative at each point of an interval, there is thus defined a new function$f'(x)\$ by the formula

\begin{equation*}
f'(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}.\tag{a}
\end{equation*}

• Other notations for the derivatives of $$f$$ at $$x=x_{0}$$ are $\bbox[#F2F2F2,5px,border:2px solid black]{\dfrac{df}{dx}(x_{0}),\quad D_{x}f(x_{0}),\quad y'(x_{0}),\quad \left.\dfrac{dy}{dx}\right|_{x=x_{0}}, \quad\left.\dfrac{d}{dx}y\right|_{x=x_{0}}.}$

• Note that in formula (a), $$x$$ is a fixed number, and $$\Delta x$$ is the variable and is independent of $$x$$. To avoid any confusion and emphasize on the independence of $$x$$ and $$\Delta x$$, it is sometimes better to show the increment of $$x$$ by a totally different letter, say $$h$$. Therefore, we write
$\bbox[#F2F2F2,5px,border:2px solid black]{f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.}\tag{b}$

• Another form of the derivative can be obtained if we write $$u=x+\Delta x$$. Then $$\Delta x=u-x$$. In this case, $$\Delta x\to0$$ if and only if $$u\to x$$, and
$\bbox[#F2F2F2,5px,border:2px solid black]{f'(x)=\lim_{u\to x}\frac{f(u)-f(x)}{u-x}.}\tag{c}$

The process of finding the derivative is called differentiation and we are said to differentiate $$y$$ with respect to $$x$$. The process of differentiation involves the following four steps:

1. Assume, at pleasure, $$\Delta x$$. Then, in the function replace $$x$$ by $$x+\Delta x$$, giving a new value of the dependent variable $$y_{1}=y+\Delta y$$.

2. Subtract the given value of $$y$$ from the new value of $$y$$, to find $$\Delta y$$.

3. Divide $$\Delta y$$ (the increment of the dependent variable) by $$\Delta x$$ (the increment of the independent variable).

4. Find the limit of $$\Delta y/\Delta x$$ when $$\Delta x$$ varies and approaches zero ($$\Delta x\to0$$).

Example 4

Given $$y=3x^{2}+5$$, differentiate $$y$$ with respect to $$x$$.

Solution

Let $$x_{0}$$ be definite value of $$x$$, and $$y_{0}=3x_{0}^{2}+5$$.
Step 1: Take $$\Delta x$$. Then $$x_{1}=x_{0}+\Delta x.$$ \begin{aligned} y_{1} & =3x_{1}^{2}+5\\ & =3(x_{0}+\Delta x)^{2}+5\\ & =3x_{0}^{2}+6x_{0}\Delta x+3(\Delta x)^{2}+5.\end{aligned} Step 2: \begin{aligned} \Delta y & =y_{1}-y_{0}\\ & =(3x_{0}^{2}+6x_{0}\Delta x+3(\Delta x)^{2}+5)-(3x_{0}^{2}+5)\\ & =6x_{0}\Delta x+3(\Delta x)^{2}.\end{aligned} Step 3: $\frac{\Delta y}{\Delta x}=6x_{0}+3\Delta x$ Step 4: $\lim_{\Delta x\to0}\frac{\Delta y}{\Delta x}=6x_{0}.$ This is the value of the derivative when $$x=x_{0}$$ (i.e. $$\left.\dfrac{dy}{dx}\right|_{x=x_{0}}$$). But $$x_{0}$$ can be any value of $$x$$; so we may drop the subscript 0 and write as a general formula $\frac{dy}{dx}=6x.$

Example 5

Find $$dy/dx$$ when $$y=1/x$$.

Solution

Let $$x_{0}$$ be definite value of $$x$$, and $$y_{0}=\dfrac{1}{x_{0}}$$ the corresponding value of $$y$$.
Step 1: Take $$\Delta x$$. Then $$x_{1}=x_{0}+\Delta x$$. $y_{1}=\dfrac{1}{x_{1}}=\dfrac{1}{x_{0}+\Delta x}$ Step 2: $\Delta y=y_{1}-y_{0}=\dfrac{1}{x_{0}+\Delta x}-\dfrac{1}{x_{0}}=-\dfrac{\Delta x}{x_{0}^{2}+x_{0}\Delta x}.$ Step 3: By division $\frac{\Delta y}{\Delta x}=-\frac{1}{x_{0}^{2}+x_{0}\Delta x}.$ Thus because $\lim_{\Delta x\to0}\left(-\frac{1}{x_{0}^{2}+x_{0}\Delta x}\right)=-\frac{1}{x_{0}^{2}},$ this is the value of the derivative when $$x=x_{0}$$. But $$x_{0}$$ may be any value of $$x$$ (except $$x_{0}=0$$); so we may drop the subscript 0 and write as a general formula $\frac{dy}{dx}=-\frac{1}{x^{2}}.$

Example 6

Find the derivative of $$y$$ with respect to $$x$$ at $$x=0$$ if $y=5|x|-9.$

Solution

Step 1: At $$x_{0}=0$$, $$y_{0}=-9$$. For $$\Delta x>0$$, $$x_{1}=0+\Delta x>0$$, we have $y_{1}=5\left|x_{1}\right|-9=5\Delta x-9.$ For $$\Delta x<0$$, $$x_{1}=0+\Delta x<0,$$ we have $$\left|x_{1}\right|=|\Delta x|=-\Delta x$$ and $y_{1}=-5\Delta x-9.$ Step 2: $\Delta y=y_{1}-y_{0}=\begin{cases} 5\Delta x & \Delta x>0\\ -5\Delta x & \Delta x<0 \end{cases}$ Step 3: $\frac{\Delta y}{\Delta x}=\begin{cases} 5 & \Delta x>0\\ -5 & \Delta x<0 \end{cases}$ Step 4: $\lim_{\Delta x\to0^{+}}\frac{\Delta y}{\Delta x}=5$ $\lim_{\Delta x\to0^{-}}\frac{\Delta y}{\Delta x}=-5$ Because the one-sided limits are different, there is no derivative at $$x=0$$.