Differential calculus is the mathematics of change. Whenever a laboratory quantity depends on another, such as the concentration of a reactant on time, the pH of a titration mixture on the volume of titrant added, or the detector signal on the concentration of a standard, differential calculus answers the question "how fast does it change here?" The answer is the derivative of the function. Geometrically, the derivative at a point is the gradient of the tangent to the graph at that point. Numerically, it is the instantaneous rate of change: the rate of a reaction at one moment, the sensitivity of a calibration curve at one concentration, or the steepness of a titration curve at one volume.
A technician meets derivatives in three ways: reading them from a graph as the gradient of a tangent, calculating them from a formula using the rules in this topic, and interpreting them with their units. A derivative always carries units of "output unit per input unit", for example mol/L per minute or pH units per mL. This topic sets out the notation for functions and derivatives, the standard derivatives that every other method builds on, and the idea of a differential equation, which links a quantity to its own rate of change.
A function is a rule that gives exactly one output for each allowed input. The input is the independent variable and the output is the dependent variable. In the equation y = f(x), x is the input, y is the output, and f is the name of the rule. The letters are chosen to suit the subject: c(t) for a concentration that depends on time, V(m) for a volume that depends on mass, A(c) for an absorbance that depends on concentration.
Reading f(x) = 3x² − 2x + 5 aloud gives "f of x equals three x squared minus two x plus five". To evaluate the function at a value, replace every x by that value, in brackets, and calculate. The value f(2) is the output when the input is 2. An expression such as f(a + 1) means that every x is replaced by the whole expression a + 1, and it is then expanded and simplified.
The derivative of y = f(x) with respect to x has two common notations, and both mean the same thing.
| Notation | Read as | Used when |
|---|---|---|
| dy/dx | dee y by dee x | The variables are named and their units matter (Leibniz notation) |
| f′(x) | f dash of x | The function is named and its value is needed at a point (Lagrange notation) |
| dc/dt | dee c by dee t | A concentration changing with time |
| d/dx | dee by dee x | An instruction to differentiate whatever follows |
The symbol dy/dx is one symbol for the derivative and is not a fraction that can be cancelled. The notation f′(2) means "the derivative of f evaluated at x = 2", which is the gradient of the tangent at x = 2. Higher derivatives use the same ideas and are met in Derivatives of higher order functions.
Example 1: The mass of crystals deposited in an evaporating dish is m(t) = 0.5t² + 2t grams, where t is the time in hours. Find m(0), m(3) and m(4), and the mass deposited between t = 3 and t = 4.
m(0) = 0.5(0)² + 2(0) = 0 g
m(3) = 0.5(9) + 2(3) = 4.5 + 6 = 10.5 g
m(4) = 0.5(16) + 2(4) = 8 + 8 = 16 g
Mass deposited between t = 3 and t = 4 = m(4) − m(3) = 16 − 10.5 = 5.5 g
Example 2: For f(x) = x² − 4x + 1, find f(2) and f(a + 1).
f(2) = 2² − 4(2) + 1 = 4 − 8 + 1 = −3
f(a + 1) = (a + 1)² − 4(a + 1) + 1
= a² + 2a + 1 − 4a − 4 + 1
= a² − 2a − 2
Check with a = 1: f(2) = 1 − 2 − 2 = −3, which agrees with the first line.
Example 3: The concentration of a reagent in a reaction vessel is c(t) mol/L, where t is the time in minutes. Explain what c′(5) = −0.012 means.
The derivative c′(5) is the rate of change of concentration at t = 5 minutes. Its units are mol/L per minute. The value −0.012 means that, at that moment, the concentration is falling at 0.012 mol/L every minute. The negative sign shows a decrease.
Standard differentiation is the use of a short list of known derivatives, together with three simple rules, to differentiate most functions met in laboratory mathematics without going back to the definition. The definition itself is developed in Differentiation by first principle.
| Function | Derivative |
|---|---|
| k (a constant) | 0 |
| xⁿ (any power n) | n xⁿ⁻¹ |
| sin ax | a cos ax |
| cos ax | −a sin ax |
| e^(ax) | a e^(ax) |
| ln x | 1/x |
For sin and cos the angle x is in radians. In the table, a and k are constants and n is any number, positive, negative or fractional.
Example 1: The temperature of a heated water bath is T(t) = 25 + 3t − 0.02t² °C, where t is in minutes. Find the heating rate at t = 20 minutes.
dT/dt = 0 + 3 − 0.02(2t) = 3 − 0.04t
At t = 20: dT/dt = 3 − 0.04(20) = 3 − 0.8 = 2.2
Answer: the bath is warming at 2.2 °C per minute.
Example 2: Differentiate y = 4x³ − 6x² + 7x − 9 and find the gradient at x = 2.
dy/dx = 4(3x²) − 6(2x) + 7 − 0 = 12x² − 12x + 7
At x = 2: 12(4) − 12(2) + 7 = 48 − 24 + 7 = 31
Answer: dy/dx = 12x² − 12x + 7 and the gradient at x = 2 is 31.
Example 3: Differentiate y = 5√x + 3/x² − 2 and find the gradient at x = 4.
Rewrite: y = 5x^(1/2) + 3x⁻² − 2
dy/dx = 5(1/2)x^(−1/2) + 3(−2)x⁻³ − 0 = (5/2)x^(−1/2) − 6x⁻³
= 5/(2√x) − 6/x³
At x = 4: 5/(2 × 2) − 6/64 = 1.25 − 0.09375 = 1.15625
Answer: dy/dx = 5/(2√x) − 6/x³ and the gradient at x = 4 is 1.15625.
Example 4: A reactant concentration is c(t) = 0.80 e^(−0.05t) mol/L, where t is in minutes. Find the rate of change at t = 10 minutes.
dc/dt = 0.80 × (−0.05) e^(−0.05t) = −0.04 e^(−0.05t)
At t = 10: −0.04 e^(−0.5) = −0.04 × 0.60653 = −0.024261
Answer: the concentration is falling at about 0.0243 mol/L per minute.
Example 5: Differentiate y = 3 sin 2x + 2 e^(0.5x) − 4 ln x and find the value of dy/dx at x = 1 (angles in radians).
dy/dx = 3(2 cos 2x) + 2(0.5)e^(0.5x) − 4(1/x) = 6 cos 2x + e^(0.5x) − 4/x
At x = 1: 6 cos 2 + e^0.5 − 4 = 6(−0.41615) + 1.64872 − 4
= −2.49688 + 1.64872 − 4 = −4.84816
Answer: dy/dx = 6 cos 2x + e^(0.5x) − 4/x, which is −4.848 at x = 1 (to 3 decimal places).
A differential equation is an equation that contains a derivative. It describes how a quantity changes in terms of the quantity itself or of the variable it depends on. In a laboratory, the rate laws of chemical kinetics are differential equations: "the rate of fall of concentration is proportional to the concentration" is written dc/dt = −kc, where k is a positive constant.
The order is the highest derivative that appears in the equation. A first-order equation contains dy/dx only, a second-order equation contains d²y/dx², and so on. The equation dc/dt = −kc is first order. The equation d²y/dt² + 9y = 0 is second order.
A solution is a function that satisfies the equation when it and its derivatives are substituted. The solution of a first-order equation contains one arbitrary constant, and the solution of a second-order equation contains two. This is the general solution, and it describes a whole family of curves. A particular solution is found by using given conditions, such as the starting concentration at t = 0, to fix the constants.
To check that a function is a solution, differentiate it as many times as the equation needs, substitute into the equation, and show that both sides are equal.
When the equation has the form dy/dx = f(x), the solution is the function whose derivative is f(x). It is found by reversing differentiation: for a term xⁿ, increase the power by 1 and divide by the new power, then add a constant C. The method is developed in Standard integration.
Example 1: State the order of (a) dc/dt = −0.3c, (b) d²y/dx² − 4 dy/dx + 3y = 0, and (c) dy/dx = x² + 1.
(a) first order, because the highest derivative is dc/dt
(b) second order, because the highest derivative is d²y/dx²
(c) first order, because the highest derivative is dy/dx
Example 2: Show that c = 0.60 e^(−0.04t) satisfies dc/dt = −0.04c, and that it gives c = 0.60 mol/L at t = 0.
Differentiate: dc/dt = 0.60 × (−0.04) e^(−0.04t) = −0.04 × (0.60 e^(−0.04t)) = −0.04c
The left side equals the right side, so the function is a solution.
At t = 0: c = 0.60 e⁰ = 0.60 × 1 = 0.60 mol/L, which meets the condition.
Example 3: A reaction whose rate does not depend on concentration (zero order) follows dc/dt = −0.02 mol/L per minute, with c = 0.50 mol/L at t = 0. Find c(t) and the time at which c = 0.20 mol/L.
The solution of dc/dt = −0.02 is c = −0.02t + C.
At t = 0: 0.50 = 0 + C, so C = 0.50.
Particular solution: c = 0.50 − 0.02t
For c = 0.20: 0.20 = 0.50 − 0.02t, so 0.02t = 0.30 and t = 15
Answer: c = 0.50 − 0.02t, and the concentration reaches 0.20 mol/L after 15 minutes.
Example 4: Find the particular solution of dy/dx = 9x² + 2x − 1 given that y = 4 when x = 1.
Reverse the power rule on each term: y = 3x³ + x² − x + C
At x = 1: 3 + 1 − 1 + C = 4, so 3 + C = 4 and C = 1
Answer: y = 3x³ + x² − x + 1
Check: dy/dx = 9x² + 2x − 1, and at x = 1, y = 3 + 1 − 1 + 1 = 4.
Example 5: The displacement of a sample tube clamped in a mechanical shaker is y = 2 sin 3t + cos 3t centimetres. Show that it satisfies d²y/dt² = −9y.
First derivative: dy/dt = 6 cos 3t − 3 sin 3t
Second derivative: d²y/dt² = −18 sin 3t − 9 cos 3t = −9(2 sin 3t + cos 3t) = −9y
Answer: the equation is satisfied.
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