Appendix G: Calculus Basics

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G.1 Calculus Basics

Here, we provide a brief and pragmatic survey of some useful well-known calculus mechanics and rules. A good reference on calculus is Flanders et al. (1970).

G.1.1 Differentiation Product Rule

$\frac{d (f g)}{d x} = f \frac{d g}{d x} + g \frac{d f}{d x}$ $\frac{d (f g)}{d x} = f \frac{d g}{d x} + g \frac{d f}{d x}$

(G.1)

Example: Differentiate the product xe^ikx:

$\frac{d (x e^{i k x})}{d x} = x (i k) e^{i k x} + e^{i k x} = e^{i k x} (1 + i k x)$ $\frac{d (x e^{i k x})}{d x} = x (i k) e^{i k x} + e^{i k x} = e^{i k x} (1 + i k x)$

G.1.2 Differentiation Quotient Rule

$\frac{d (f / g)}{d x} = g^{- 2} (g \frac{d f}{d x} - f \frac{d g}{d x})$ $\frac{d (f / g)}{d x} = g^{- 2} (g \frac{d f}{d x} - f \frac{d g}{d x})$

(G.2)

G.1.3 Differentiation Power Rule

If n is an integer,

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

(G.3)

Example: Differentiate (x² + 1)²:

$\frac{d {(x^{2} + 1)}^{2}}{d x} = 4 x (x^{2} + 1)$ $\frac{d {(x^{2} + 1)}^{2}}{d x} = 4 x (x^{2} + 1)$

G.1.4 Differentiation Chain Rule

If y and x are functions of t,

$\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}$ $\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t}$

(G.4)

Example: Differentiate the function $y = e^{t^{2} + 2 t + 1}$ $y = e^{t^{2} + 2 t + 1}$ Set y = e^x and x = t² + 2t + 1. Then apply the chain rule:

$\frac{d y}{d t} = e^{x} (2 t + 2) = e^{t^{2} + 2 t + 1} (2 t + 2)$ $\frac{d y}{d t} = e^{x} (2 t + 2) = e^{t^{2} + 2 t + 1} (2 t + 2)$

G.1.5 Integration by Parts

$\int f d g = f g - \int g d f$ $\int f d g = f g - \int g d f$

(G.5)

Example: Integrate by parts $\int x e^{i k x} d x$ $\int x e^{i k x} d x$ . Set f = x, df = dx, dg = e^ikxdx, and g = e^ikx/ik.

Then apply Equation G.5:

$\int x e^{i k x} d x = \frac{x}{i k} e^{i k x} - \int \frac{e^{i k x}}{i k} d x = \frac{e^{i k x}}{i k} (x - \frac{1}{i k}) + C$ $\int x e^{i k x} d x = \frac{x}{i k} e^{i k x} - \int \frac{e^{i k x}}{i k} d x = \frac{e^{i k x}}{i k} (x - \frac{1}{i k}) + C$

where C is a constant. Differentiation of F(x) = (e^ikx/ik)(x − (1/ik)) + C, using the product rule leads back to xe^ikx.

Reference

Flanders, H., Korfhage, R. R., and Price, J. J. (1970). Calculus. Academic Press, New York.

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Table of Contents for Appendix G: Calculus Basics

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Appendix G: Calculus Basics