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1995-03-13
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3KB
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DIFFERENTIAL CALCULUS
A function f(x) is continuous at x = a, if and only if
lim f(x) = f(a).
x─>a
Limits of continuous functions can be evaluated by substitution. If a O/O
or ∞/∞ form results then algebraic simplification must precede
substitution.
Given that lim f(x) and lim g(x) exist, the following are true:
x->a x->a
1. lim (f(x) ± g(x)) = lim f(x) ± lim g(x)
x->a x->a x->a
2. lim (f(x) ∙ g(x)) = lim f(x) ∙ lim g(x)
x->a x->a x->a
3. lim (f(x)/g(x)) = lim f(x)/lim g(x) provided lim g(x) ╪ O
x->a x->a x->a x->a
4. lim f(x) = L <=> lim f(x) = L and lim f(x) = L .
x->a x->a+ x->a-
The derivative of f(x) with respect to x evaluated at x is
°
df(x ) f(x + h) - f(x ) f(x) - f(x )
° ° ° °
─────── = lim ──────────────── = lim ────────────
dx h->O h x->x x - x
° °
if the limit exists.
Geometrically, the derivative gives the slope of the curve at the point
where it is evaluated.
Rules for differentiation you will need are:
d ┌ ┐ df(x) dg(x)
1. ── │f(x) ± g(x)│ = ───── ± ─────
dx └ ┘ dx dx
d ┌ ┐ dg(x) df(x)
2. ── │f(x) ∙ g(x)│ = f(x) ───── + g(x) ─────
dx └ ┘ dx dx
df(x) dg(x)
┌ ┐ g(x) ───── - f(x) ─────
d │f(x)│ dx dx
3. ── │────│ = ─────────────────────────
dx │g(x)│ ┌ ┐2
└ ┘ │g(x)│
└ ┘
n
dx n-1 d sin x
4. ─── = nx 5. ─────── = cos x
dx dx
d cos x d tan x 2
6. ─────── = -sin x 7. ─────── = sec x
dx dx
d cot x 2 d sec x
8. ─────── = -csc x 9. ─────── = (sec x)(tan x)
dx dx
d csc x
1O. ─────── = -(csc x)(cot x)
dx
The chain rule gives the derivative of f(u(x)) with respect to x as:
d f(u(x)) d f(u) du
─────── = ─────── ∙ ── .
dx du dx
This procedure can be used on formula 4-1O above to produce additional
formula.
The expression y = f(x). is called explicit. Expressions which are
not explicit are called implicit, e.g. xy = 3. To differentiate implicit
expressions, differentiate both sides of the expression with respect to the
independent variable. Use the chain rule where needed.