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Basic Differentiation

MATH 2511, BCIT

Technical Mathematics for Geomatics

January 16, 2023

Introduction to Calculus

Calculus solves many problems for which it was not originally designed. The initial motivation for calculus was to find the slope of a tangent on a curve and the area of a region bounded by a curve.

A Real-Life Example II

It appears to make sense only if we calculate the velocity given an interval of time rather than just one point in time. For example, the velocity between t = 2 and t = 3 is

v[2,3] =

f (3) − f (2) 3 − 2

More generally,

v[2,t] = g (t) = f (t) − f (2) t − 2

4(t^2 − 4) t − 2

g is not defined at t = 2, but it is defined all around t = 2, so we can ask ourselves the question: what happens when t → 2 from below; or when t → 2 from above? It turns out that either way, the number approaches 16.

Velocity at a Point

If we found the limit as t → 2 of

v[2,t] = g (t) = f (t) − f (2) t − 2

4(t^2 − 4) t − 2 it would serve as an intuitive definition of what a velocity is at a point (instead of on an interval). Unfortunately, the limit has the indeterminate form lim t→ 2

4(t^2 − 4) t − 2

However, notice that for t 6 = 2,

g (t) =

4(t^2 − 4) t − 2

4(t + 2)(t − 2) t − 2 = 4(t + 2) (5)

Tangent Lines II

Here is an example of a secant line.

Tangent Lines III

Now imagine t 1 and t 2 moving closer and closer together at a point a (for the train, we used a = 2). If both of these limits exist and agree with each other, we have a velocity at a point,

lim t→a v[a,t] for t > a (8)

tlim→a v[t,a]^ for^ t^ <^ a^ (9) This velocity at a point is also the slope of the line that just touches the function graph without crossing it. We call it a tangent line at t = a. The slope of the tangent line is sometimes also called the rate of change.

Derivatives

The derivative of a function f with respect to x is the function f ′ (read “f prime”),

f ′(x) = lim h→ 0

f (x + h) − f (x) h

The domain of f ′^ is the set of all x where the limit exists.

Derivatives Diagram I

Basic Rules of Differentiation

Rule 1 Derivative of a Constant

f ′(x) = 0 for f (x) = c (12) Reason:

f ′(x) = lim h→ 0

f (x + h) − f (x) h

= lim h→ 0

c − c h

= lim h→ 0

Basic Rules of Differentiation

Rule 2 The Power Rule

f ′(x) = nxn−^1 for f (x) = xn^ (14) This rule applies for any n ∈ R. The proof is messy. However, we can show how the rule is justified for n = 2 and n = 12.

Basic Rules of Differentiation

Case 2: n = (^12)

f ′(x) = lim h→ 0

f (x + h) − f (x) h

= lim h→ 0

x + h −

x h

lim h→ 0

x + h −

x) · (

x + h +

x) h · (

x + h +

x)

= lim h→ 0

x + h − x h · (

x + h +

x) Using the One Disagreement Rule, this equals

lim h→ 0

x + h +

x

x +

x

x−^

(^12) (16)

Basic Rules of Differentiation

Rule 3 Derivative of a Constant Multiple of a Function

g ′(x) = c · f ′(x) for g (x) = c · f (x) (17) Reason:

g ′(x) = lim h→ 0

g (x + h) − g (x) h = lim h→ 0

c · f (x + h) − c · f (x) h

lim h→ 0 c ·

f (x + h) − f (x) h = c · lim h→ 0

f (x + h) − f (x) h = c · f ′(x) (18)

Basic Differentiation Exercises

Exercise 1: Find the derivatives for the following functions. (^1) f (x) = 4x^5 + 3x^4 − 8 x^2 + x + 3 (^2) f (t) = t 52 + (^) t^53 (^3) g (z) = 2z − 5 √z

Exercise 2: Find the slope and an equation of the tangent line to the graph of f (x) = 2x + (1/

x) at the point (1, 3).

Basic Differentiation Exercises

Exercise 3: Find the derivative for the following function.

f (x) = 5x

4 (^3) − 2 3

x

3 (^2) + x^2 − 3 x + 1 (21)