[Calculus] III. Derivatives

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Derivatives

Definition 3.1.1.  The derivative of a function $f$ at a number $a$, denoted by $f'(a)$, is

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

If $x = a + h$, then

$\displaystyle f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}$

Definition 3.1.2.  If we replace $a$ by a variable $x$ in Definition 3.1.1, we obtain

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

Given any number $x$ for which this limit exists, we assign to $x$ the number $f'(x)$. So we can regard $f'$ as a new function, called the derivative of $f$.

Remark 3.1.3.  If we use the traditional notation $y = f(x)$ to indicate that the independent variable is $x$ and the dependent variable is $y$, then some common alternative notations for the derivative are as follows:

$\displaystyle f'(x) = y' = {\operatorname{d}\!y\over\operatorname{d}\!x} = {\operatorname{d}\!f\over\operatorname{d}\!x} = {\operatorname{d}\over\operatorname{d}\!x}f(x) = Df(x) = D_{x}f(x)$

The symbols $D$ and $\displaystyle {\operatorname{d}\over\operatorname{d}\!x}$ are called differentiation operators because they indicate the operation of differentiation, which is the process of calculating a derivative. The symbol $\displaystyle {\operatorname{d}\!y\over\operatorname{d}\!x}$, which was introduced by Leibniz, should not be regarded as a ratio (for the time being); it is simply a synonym for $f'(x)$.

If we want to indicate the value of a derivative $\displaystyle {\operatorname{d}\!y\over\operatorname{d}\!x}$ at a specific number $a$, we use the notation

$\displaystyle {\operatorname{d}\!y\over\operatorname{d}\!x} \Bigg\vert_{\ x = a} \qquad \text{or} \qquad {\operatorname{d}\!y\over\operatorname{d}\!x} \Bigg]_{x = a}$

which is a synonym for $f'(a)$.

Example 3.1.4.  If $f(x) = x^{3} - x$, find a formula for $f'(x)$.

Solution.

$\displaystyle \begin{align} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} = \lim_{h \to 0}\frac{[(x + h)^{3} - (x + h)] - [x^{3} - x]}{h} \\[7pt] &= \lim_{h \to 0}\frac{x^{3} + 3x^{2}h + 3xh^{2} + h^{3} - x - h - x^{3} + x}{h} \\[7pt] &= \lim_{h \to 0}\frac{3x^{2}h + 3xh^{2} + h^{3} - h}{h} \\[7pt] &= \lim_{h \to 0}(3x^{2} + 3xh + h^{2} - 1) = 3x^{2} - 1 \end{align}$

$\Box$

Theorem 3.1.5.  The tangent line to the curve $y = f(x)$ at the point $P(a, f(a))$ is the line through $P$ with slope

$\displaystyle m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

provided that this limit exists. And if $x = a + h$, the slope of the tangent line is

$\displaystyle m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

Figure 1

Remark 3.1.6.  We can now say that the tangent line to $y = f(x)$ at $(a, f(a))$ is the line through $(a, f(a))$ whose slope is equal to $f'(a)$, the derivative of $f$ at $a$. In other words,

$\displaystyle m = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = f'(a)$

Therefore, if we use the point-slope form of the equation of a line, we can write an equation of the tangent line to the curve $y = f(x)$ at the point $(a, f(a))$:

$\displaystyle y - f(a) = f'(a)(x - a)$

Definition 3.1.7.  A function $f$ is differentiable at $a$ if $\displaystyle f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}$ exists. It is differentiable on an open interval $(a,\ b)$ if it is differentiable at every number in the interval.

Theorem 3.1.8.  If $f$ is differentiable at $a$, then $f$ is continuous at $a$.

Proof.  To prove that $f$ is continuous at $a$, we have to show that $\displaystyle \lim_{x \to a}f(x) = f(a)$. We do this by showing that the difference $f(x) - f(a)$ approaches $0$.

The given information is that $f$ is differentiable at $a$, that is,

$\displaystyle f'(a) = \lim_{x \to a}\frac{f(x) - f(a)}{x - a}$

exists. To connect the given and the unknown, we divide and multiply $f(x) - f(a)$ by $x - a$ (which we can do when $x \neq a$) :

$\displaystyle f(x) - f(a) = \frac{f(x) - f(a)}{x - a}(x - a)$

Thus, we can write

$\displaystyle \begin{align} \lim_{x \to a}\left[ f(x) - f(a) \right] &= \lim_{x \to a}\frac{f(x) - f(a)}{x - a}(x - a) \\[7pt] &= \lim_{x \to a}\frac{f(x) - f(a)}{x - a} \cdot \lim_{x \to a}(x - a) \\[7pt] &= f'(a) \cdot 0 = 0 \end{align}$

Therefore,

$\displaystyle \begin{align} \lim_{x \to a}f(x) &= \lim_{x \to a}\left[ f(a) + f(x) - f(a) \right] \\[7pt] &= \lim_{x \to a}f(a) + \lim_{x \to a}\left[ f(x) - f(a) \right] \\[7pt] &= f(a) + 0 = f(a) \end{align}$

Thus, $f$ is continuous at $a$.

$\Box$

Definition 3.1.9.  The second derivative of $y = f(x)$ is

$\displaystyle f''(x) = {\operatorname{d}\over\operatorname{d}\!x} \left( {\operatorname{d}\!y\over\operatorname{d}\!x} \right) = {\operatorname{d}^{2}\!y\over\operatorname{d}\!x^{2}}$

The third derivative of $y = f(x)$ is

$\displaystyle f'''(x) = {\operatorname{d}\over\operatorname{d}\!x} \left( {\operatorname{d}^{2}\!y\over\operatorname{d}\!x^{2}} \right) = {\operatorname{d}^{3}\!y\over\operatorname{d}\!x^{3}}$

$\begin{matrix} \vdots \\[7pt] \end{matrix}$

The $n$th derivative of $y = f(x)$ is

$\displaystyle f^{(n)}(x) = {\operatorname{d}^{n}\!y\over\operatorname{d}\!x^{n}}$

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Rates of Change

Definition 3.2.1.  Suppose $y$ is a quantity that depends on another quantity $x$. Thus $y$ is a function of $x$ and we write $y = f(x)$. If $x$ changes from $x_1$ to $x_2$, then the change in $x$ (also called the increment of $x$) is

$\Delta x = x_{2} - x_{1}$

and the corresponding change in $y$ is

$\Delta y = f(x_{2}) - f(x_{1})$

Definition 3.2.2.  The difference quotient

$\displaystyle \frac{\Delta y}{\Delta x} = \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}}$

is called the average rate of change of $y$ with respect to $x$ over the interval $\left[ x_{1},\ x_{2} \right]$. And the limit of the average rate of change

$\displaystyle \lim_{\Delta x \to 0}\! \frac{\Delta y}{\Delta x} = \lim_{x_{2} \to x_{1}}\! \frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}}$

is called the (instantaneous) rate of change of $y$ with respect to $x$ at $x = x_{1}$.

Definition 3.2.3.  If $s = f(t)$ is the position function of a moving object, the average velocity of the object from $t = t_{0}$ to $t = t_{0} + \Delta t$ is

$\displaystyle v_{\text{av}} = \frac{s(t_{0} + \Delta t) - s(t_{0})}{\Delta t}$

and the velocity (or instantaneous velocity) at time $t = t_{0}$ is

$\displaystyle v(t_{0}) = {\operatorname{d}\!s\over\operatorname{d}\!t} \Bigg\vert_{\ t = t_{0}} = \lim_{\Delta t \to 0}\!\frac{s(t_{0} + \Delta t) - s(t_{0})}{\Delta t}$

and the acceleration at time $t = t_{0}$ is

$\displaystyle a(t_{0}) = {\operatorname{d}^{2}\!s\over\operatorname{d}\!t^{2}} \Bigg\vert_{\ t = t_{0}} = \lim_{\Delta t \to 0}\!\frac{v(t_{0} + \Delta t) - v(t_{0})}{\Delta t}$

and the jerk at time $t = t_{0}$ is

$\displaystyle j(t_{0}) = {\operatorname{d}^{3}\!s\over\operatorname{d}\!t^{3}} \Bigg\vert_{\ t = t_{0}} = \lim_{\Delta t \to 0}\!\frac{a(t_{0} + \Delta t) - a(t_{0})}{\Delta t}$

Example 3.2.4.  The position of a particle is given by the equation

$\displaystyle s = f(t) = t^{3} - 6t^{2} + 9t$

where $t$ is measured in seconds and $s$ in meters. Find the velocity at time $t$.

Solution.  The velocity function is the derivative of the position function.

$\displaystyle v(t) = {\operatorname{d}\!s\over\operatorname{d}\!t} = 3t^{2} - 12t + 9$

$\Box$

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The body text and the figures were excerpted from James Stewart, Calculus 8E.