Derivatives - Essential Mathematics for Chemical Engineers

Tangent line and rate of change¶

A tangent line touches a curve exactly once.

$Graph of tangent line as slope of curve$

How can we estimate the slope of the line at $x_0$ ? First, evaluate $f(x_0)$ . Then, evaluate $f(x_1)$ at another point nearby. The slope is:

\frac{f(x_1)-f(x_0)}{x_1-x_0}

(1)

If we make $x_1$ very close to $x_0$ , we will get the slope of the tangent line. This suggests uses of a limit! We call the slope of the tangent to f at $x$ the derivative of f.

The derivative represents how fast the function is changing (rate of change). This is especially useful in physics when the function represents a coordinate, and the independent variable is time.

Example: Ball moving with constant velocity¶

A ball has position coordinate x and is moving with constant velocity v, so

x(t) = x_0 + vt

(3)

where $x_0$ is the initial position that it starts from.

$Ball with constant velocity example$

Show that the first derivative of the position x is the velocity v.

\begin{align} x'(t) &= \lim_{h\to 0} \frac{x(t+h)-x(t)}{h} \\ &= \lim_{h\to 0} \frac{[x_0+v(t+h)]-[x_0+vt]}{h} \\ &=\lim_{h\to 0} \frac{vh}{h} \\ &= v \end{align}

(4)

Formal definition¶

The derivative is also a function!

$Function and derivative$

For example, if $f(x) = x^2$ ,

\begin{align} f'(x) &= \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \\ &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \\ &= \lim_{h \to 0} \frac{2xh + h^2}{h} \\ &= \lim_{h \to 0} (2x + h) \\ &= 2x \end{align}

(5)

This function represents how the slope of the line tangent to f is changing with x. For this function, the slope continually increases! Some additional examples:

Differentiability¶

To be differentiable, this limit must exist. A function would not be differentiable at a jump, cusp, coner, or vertical tangent

$Function and derivative$

Higher-order derivatives¶

We can try to take the derivative of any function, including the derivative itself! For example, consider $f(x) = x^2$ , which has $f'(x) = 2x$ . We can evaluate the derivative of $f'$ , which we will denote $f''$ using the same formal definition:

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

(8)

We call $f''$ the second derivative of f. Repeating the process we can obtain even higher-order derivatives. These may be denoted by additional primes (e.g., $f'''$ ), but often by a superscript (e.g., $f^{(3)}$ ) since it gets clunky to count too many primes.

Higher-order derivatives tell us about how lower-order derivatives are changing. For example, the second derivative tells us how the first derivative changes.

In a physical setting, we may have names for these different derivatives. For example, if we have the position x, the velocity v is the first derivative of x, while the acceleration a is the first derivative of v and second derivative of x.

Position	Velocity	Acceleration
x	x’	x’’
	v	v’
		a

Example: Ballistic motion¶

Consider the two-dimensional ballistic motion:

$Ballistic motion$

The equations of motion for the horizontal position x and vertical position y are:

\begin{align} x(t) &= u_0 t & \\ y(t) &= v_0 t - \frac{1}{2} g t^2 \end{align}

(9)

where $u_0$ and $v_0$ are the initial x and y components of the velocity, and g is the acceleration due to gravity. Find the velocity and acceleration in x and y.

In the x direction, call the velocity u. Then,

u = x' = u_0

(10)

Since u is a constant, the first derivative of u (second derivative of x) must be zero, and there is no acceleration in the x direction.

In the y direction, call the velocity v. Then,

\begin{align} v &= y' = v_0 - gt \\ a &= v' = y'' = -g \end{align}

(11)

The velocity in y is constantly increasing from the constant acceleration a due to gravity. This makes sense from what we know from physics!