Unit IV Derivatives 20 Hours Finish by Christmas
Calculus There two main streams of Calculus: Differentiation Integration
Differentiation is used to find the rate of change of variables relative to one another. i.e. finding the slope of tangents lines Differentiation is also called taking the derivative of a function. Integration is used to find the accumulation of one quantity over time. i.e. Area under a curve
d From PHYSICS in Science 1206 Tangents to a displacement vs time curve gives the instantaneous velocity at a point What is the velocity at 3 seconds? t This is Differential Calculus
v 4 3 What did the area under a velocity time graph give you? 2 Displacement 1 1 2 3 t What is the displacement after 3 seconds? This is Integral Calculus
In this unit of Derivatives we will find: Concept of the derivative. Derivative at a point. Derivative as a function. Computations of Derivative using definition of derivative Computations of Derivative using differentiation rules. Implicit Differentiation
Part A Introduction to Differential Calculus Demonstrate an understanding of the concept of a derivative and evaluate derivatives of functions using the definition of derivative.
The Details Explain what is meant by the term tangent. Calculate average rate of change. Identify the instantaneous rate of change of a function at a point as the limiting value of a sequence of average rates of change.
Derivatives: Tangents and Velocity First application of Calculus that we look at. What is meant by a tangent? Touches a curve at one point Circle Curve
Tangents in the Real World
For tangents we consider the line to touch at only a local level. We do not worry about cases like this Tangent here
Why are tangents important in Mathematics? A tangent gives an indication of the steepness of a curve at a given point. The slope of the tangent at a point gives the instantaneous rate of change of the function at that point. Consider the graph on the next page.
N P M The curve at M is steeper than the curve at P The curve is decreasing at N as compared to increasing at P and M We can get this information, without drawing the graph, by looking at the slopes of the tangent lines at these points.
Q R What is the local max point on this graph? Q What is the local min point on this graph? R What is the slope of the tangent lines at these extreme points? ZERO
Rate of Change
Definition: Rate of Change: The change in one quantity compared with the change in another quantity. Examples: Change in distance compared to time Speed Change in displacement compared to time Velocity Change in velocity compared to time Acceleration Change in temperature compared to altitude. Adiabatic lapse rate 2 o C 1000ft
Example: Lets look at a trip from Massey Drive to Bonne Bay Pond Time Location Total Distance 5:00 pm Massey Drive 5:10 pm Steady Brook 5:40 pm Deer Lake 0 km 10 km 50 km 5:51 pm Bonne Bay Pond 80 km What is the average speed for the complete trip? d 80km Average Speed 1hr t 51min 60min What is the average speed for each part of the trip?
Finding the average change in a function value (y) Consider the graph of y = x 2 4 Find the average change in y when x changes from 1 to 2. Find the slope of the secant line joining the 2 points y y 2 1 m x 2 x 1 0 ( 3) 2 1 3 Therefore, the average change in y as x changes from 1 to 2 is 3
Note A line that connects 2 points on a curve is called a secant line.
In general, for any function y = f(x) The Average Rate Of Change in y (f(x)) for x changing from x 1 to x 2 is found using the slope formula: y y2 y1 f ( x 2) f x1 AROC x x x x x 2 1 2 1 This formula provides the slope of the secant line between the points (x 1,f(x 1 )) and (x 2,f(x 2 )).
Example: For the function f(x) = -5x 2 +10x + 20 find the average rate of change in f(x) from x 1 = 0 to x 2 = 1
Instantaneous Rate of Change IROC How can you tell how fast you are going in a car? Look at the Speedometer What if your speedometer is broken, but your odometer still works? You could calculate your speed by finding the time it takes to go 1 km and divide 1km by that time.
How could you get a better idea how fast you are going right now, at this instant?? Find the time to go 0.1 km and divide 0.1km by that time. The idea is that by finding the average speed for smaller intervals you get a better idea at what your instantaneous speed is!
For graphs: The IROC for a graph is the SLOPE of a tangent drawn at a particular point For example: The IROC of y with respect to x at x = 1 is found by determining the slope of the tangent at the point (1,1)
Because it is difficult to accurately draw tangents we can approximate the slope of the tangent by finding the average slope of secant lines around the point in question
Here m sec > m tan Case I
Here m sec < m tan Case II
Lets consider Case I Where we take the point to the right of x = 1 Lets take a point close to x = 1, say x = 2. Draw the secant line. m sec y x f ( x ) x y x 2 1 2 1 f x 2 1 x 2 1 NOTE : f ( x ) x 2 m sec 2 1 2 1 2 2 4 1 1 3 We say that the slope of the tangent is close to 3
How can we get a better approximation? We take the point closer to x = 1 Lets use x 2 = 1.5. Draw the secant line. m sec f ( x2) f x1 x x 2 1 m sec 1.5 1 1.5 1 2 2 2.25 1 0.5 1.25 2.5 0.5 We say that the slope of the tangent is close to 2.5
Is this approximation close Not Yet! enough to the slope?
Lets usex 2 = 1.1. Draw the secant f ( x ) msec x m sec f x 2 1 x 2 1 2 2 1.1 1 1.1 1 1.21 1 0.1 0.21 0.1 2.1 We say that the slope of the tangent approximately 2.1 Or that the IROC of y with respect to x at x = 1 is 2.1
Not Yet! Is this approximation close enough to the slope? Where do we stop?? Well that is where Calculus comes in! We will find that the slope of the tangent line is the limiting value of the slopes of the secant lines as the secant line is drawn closer and closer to the tangent line.
Remember differentiation is used to find the rate of change of variables relative to one another. i.e. IROC which is represented graphically as the slope of tangents So how do we use calculus to find these slopes?
Finding slopes of tangent lines The first step to finding the slope of tangent lines is to make approximations using the slope of secant lines. Recall: A secant line is a line that connects 2 points on a curve.
Find the slope of the tangent line to the curve y = x 2 at the point P(2, 4) We find the approximate slope of the tangent by drawing lines (secant lines) that have approximately the same slope as the tangent line and calculate their slope.
Find the slope of the tangent line to the curve y = x 2 at the point P(2, 4) For the first approximation we draw line PQ with Q close to P, say (1,1). y2 y1 4 1 mpq 3 x x 2 1 2 1 Thus we can say that the slope of the tangent line is close to 3.
How do we get a better approximation for the slope? We take the point Q to be closer and closer to P y2 y1 x Y m PQ x 1 1 3 x 2 1 1.5 2.25 3.5 x is approaching 2 x 2 1.9 1.99 3.61 3.961 3.9 3.99 1.999 3.996001 3.999 Thus we can say that the slope of the tangent line is close to 4.
We can take the point Q to be closer and closer to P from the right x 3 2.5 Y m PQ y x 9 5 6.25 4.5 y x 2 1 2 1 2.1 2.01 2.001 4.41 4.0401 4.004001 4.1 4.01 4.001 Thus we can say that the slope of the tangent line is close to 4.
Thus we could say that as Q gets closer to P, the slope of the secant line gets closer to 4. From this we could guess that the slope of the tangent line at P is 4. We say that the slope of the tangent line is the limit of the slopes of the secant lines as Q approaches P.
Symbolically: lim m PQ QP m tan NOTE: m lim m PQ QP PQ Slope of secant lim y x x 2 2 1 2 1 2 x y x x Slope of tangent 4 2 lim x 2 4 x 2 x 2 x lim x 2 4 2 ( x 2)( x 2) x 2
What is the equation of the tangent line at P?
Tangent/Velocity Find the slope of a tangent line to the curve y = f(x) at point P (a, f(a)) f(x) f(a) a P x Q m m PQ PQ y x y x 2 1 2 1 f ( x ) f ( a) x a And as x gets closer to a, the secant line gets closer to the tangent line.
In fact, the tangent line is the limiting line for the secant line as x gets closer to a. m lim m tan PQ QP m tan f ( x ) f ( a) lim xa x a
Definition: The tangent line to the curve y = f(x) at the point P(a, f(a)) is the line through P with slope f ( x ) f ( a) m lim xa x a provided the limit exists!
Examples 1. Find the equation of the tangent line to the curve y = x 2 at the point (- 1, 1). Use f ( x ) f ( a) the formula mtan lim. xa x a In this problem a = - 1 and f(x) = x 2.
2.Find the equation of the normal line to the graph y = x 2 + x at x = 1. NOTE: The normal line is perpendicular to the tangent. Find the slope of the tangent line first using the formula and then find the perpendicular slope. Use this to find the equation of the normal line.
3.Find the slope of the tangent to the 1 curve f(x) = at the point ( 0, 1) x 1
Velocity Example: What is the instantaneous rate of change (IROC) aka velocity at t = 2 s for an object traveling according the displacement function s(t) = -5t 2 +30t?
TEXT: Page 81 # 3-6. 9, 10, 11, 15, 16, 18
New Definition: The definition of the tangent line can be changed into a different form by introducing a variable, h, to represent the distance that x is away from a. Consider the graph on the next page.
m tan a x SO, as x approaches a, h approaches zero thus taking the limit as is the same as taking the limit as h 0 h x a f ( x ) f ( a) lim xa x a Also h = x - a OR x = a + h Change the limit below such that x is removed
m m tan tan f ( x ) f ( a) lim xa x a f ( a h) f ( a) lim h h0
Apply this new formula we just derived to find the slope of the tangent line at the 1 point P ( 1, 1) for the function m tan f ( a h) f ( a) lim h0 h f( x) x
Find the equation of the tangent line 3 at the point for the function 2x 1 f( x) 3x 1 m tan 1, 2 f ( a h) f ( a) lim h0 h
The Velocity Problem
The Velocity Problem Suppose an object moves along a straight line according to an equation of motion s = f(t), where s is the displacement (directed distance) of the object from the origin at time t. The function f that describes the motion is called the position function of the object.
In the time interval from t = a to t = a + h the change in position is f (a + h) f (a). The average velocity over this time interval is: average velocity displacement f ( a h ) f ( a ) time h
From the graph of position and time we can see that the average velocity over this time interval is the same as the slope of the secant line PQ.
Now suppose we compute the average velocities over shorter and shorter time intervals [a, a + h]. In other words, we let h approach 0. We have defined the velocity (or instantaneous velocity) v (a) at time t = a to be the limit of these average velocities: va ( ) lim h0 f a h f a ( ) ( ) This gives the slope of the tangent line to the graph of s = f(t) when t = a. h
Example Suppose that a ball is dropped from the upper observation deck of the CN Tower, 450 m above the ground. (a) What is the velocity of the ball after 5 seconds? Use the equation of motion s = f (t) = 4.9t 2 to find the velocity v (a) after a seconds: f ( a h) f ( a) va ( ) lim h0 h
(b) How fast is the ball traveling when it hits the ground? Since the observation deck is 450 m above the ground, the ball will hit the ground at the time t 1 when s (t 1 ) = 450 The velocity of the ball as it hits the ground is therefore
# 12,13,14a) b) Text Page 81
Definition of Derivative Check up Each of the following represents the derivative of a function, f, at some number a. State f and a in each case. 1 1 ( i ) lim x 2 x 2 x 2 ( ii )lim h 0 2 2 2h 2 h ( iii )lim h 0 9h 3 h
In the Previous Section we: Defined and evaluated the derivative at x = a : m m tan tan f ( x ) f ( a) lim xa x a f ( a h) f ( a) lim h0 h Determined the equation of the tangent line and normal line to a graph of a relation at a given point. Used the derivative to find the velocity from position (displacement) function.
In the next section we will: Define and determine the derivative of a function using f ( x h) f ( x ) f( x) lim h0 h (limited to polynomials of degree 3, square root and rational functions with linear terms).
Use alternate notation interchangeably to express derivatives Determine whether a function is differentiable at a given point. Explain why a function is not differentiable at a given point, and distinguish between corners, cusps, discontinuities, and vertical tangents. Determine all values for which a function is differentiable, given the graph.
Sketch a graph of the derivative of a function, given the graph of a function. Sketch a graph of the function, given the graph of the derivative of a function.
Part B Derivatives Slope of a tangent line to the curve y = f(x) where x = a is given by: m tan f ( a h) f ( a) lim h0 h The velocity of an object with position s(t) at a time t = a is given by; s( a h) s( a) va ( ) lim h0 h
These two are basically the same equation. In fact, this type of equation is found whenever you calculate any rate of change in science, engineering, or business. It occurs so often that it is given a special name and notation
Definition of Derivative The derivative of a function f at x is denoted by f( x) and is expressed as: f ( x h) f ( x ) f( x) lim h0 h provided the limit exists!!
Domain of the derivative function. The domain of the new function f( x) is the set of all numbers x for which the limit exists. Also the domain of f( x) is a subset of the domain of f(x) because f(x) is used in the definition of f( x)
Examples: 1. Find the derivative of the function f(x) = x 2 + 2 f ( x h) f ( x ) f( x) lim h0 h
Examples: 2. Find the derivative of the function f(x) = x 3 x 2 + 1 f ( x h) f ( x ) f( x) lim h0 h
3. Find g ( x) for g( x) lim h0 Examples: 1 gx ( ) x 2 State the domain of g(x) and g( x h) g( x ) h g( x)
Examples: ( ) x 1 4. Find g ( x) for g x State the domain of g(x) and g( x) lim h0 g( x h) g( x ) h g( x)
Examples: 5. Differentiate (Find f ( x) ) f ( x h) f ( x ) f( x) lim h0 h f( x) 1 2 x
Find s() t CRAZY ONE for st () 1 2t 3t 4
Text Page 82 #31-36 Page 93 #19, 21, 23, 25, 27
Notes on Notation The derivative of a function y = f(x) at x is: f ( x h) f ( x ) f( x) lim h0 The derivative can also be written as: f( x) y OR AS h dy df dx d f dx ( x dx ) Leibniz Notation means the derivative of y (or f) with respect to x D f ( x ) D f ( x ) x
dy dx Leibniz Notation should not be regarded as a ratio. It is just another way of saying The advantage of the Leibniz notation is that both the independent variable (x) and the dependent variable (y) are shown in the derivative Example: Velocity v ( t ) s( x ) f( x) ds dt
Evaluating a Derivative To evaluate a derivative of y = f(x) at a certain value x = a we could say: f( a) or in Leibniz notation we would say: dy dx x a
d and D x are called differentiation dx operators. Differentiation is the process of calculating a derivative A function is differentiable at x = a if f( a) exists. A function is differentiable over an interval if it is differentiable at every number in the interval.
Theorem If a function f is differentiable at x = a then f is continuous at x = a. This means differentiable continuous but continuous differentiable However not continuous not differentiable
If a function is continuous at x = a, its graph is uninterrupted at x = a. If a function is differentiable at x = a, its graph is uninterrupted and smooth at x = a. Smooth means no sharp points. Also even if the function is smooth and continuous at x = a it would not be differentiable the if the tangent was a vertical line. (Infinite slope)
Examples: Discuss the differentiability of the following: This graph is 1. y = x 2 continuous and differentiable everywhere. What is y? y This is defined everywhere. 2x
2. If a function is NOT continuous at x = a then it is NOT differentiable a Note: f(x) is differentiable for all other values other than x = a
3. Is g(x) differentiable at x = 1? NO! g(x) is NOT continuous at x = 1 thus its is NOT differentiable at x = 1
4. y = x This function is continuous for but it is not differentiable at (0,0) There is a sharp point, (or corner point) at x = 0. With no unique tangent the function is not differentiable at x = 0,
Lets find the derivative of f(x) = x at x = 0 f (0 h) f (0) 0 h 0 f (0) lim lim h0 h h0 h h lim undefined h 0 h h lim h h lim 0 h h 0 h0 h 0 h lim 1 h h lim 1 h h Therefore the derivative does not exist at x = 0
5. y = f(x) This function is continuous, but it is not differentiable at (1,3) There is a sharp point at x = 1. There are two tangent lines depending on whether you approach x= 1 from left or right With no unique tangent the function is not differentiable at x = 1
2 6. Sketch the graph of (use technology) y y x 3 This function is continuous in the interval x but it is not differentiable at (0,0) There is a cusp at x = 0 A cusp is an extreme case of a corner point where the slopes of the secant lines approach from one side of the curve and - from the other. With no unique tangent the function and the fact that the tangent has an infinite slope the function is not differentiable at x = 0,
7. y = g(x) Is g(x) differentiable at x = 1? NO! g(x) is continuous at x = 1 but it is NOT differentiable at x = 1 because f(x) has a vertical tangent at x = 1. The tangent has an undefined slope. (Infinite slope)
In general, Functions whose graphs have : corners Cusps sharp points Vertical tangents are not differentiable at these points
Text Page 94 #33-37
Differentiability Check- UP 1. Consider the function f( x) 2 x, x 1 2 x 2, x 1 A) Is the function continuous at x = 1 B) Is the function differentiable at x = 1?
Is there a cusp, vertical tangent, discontinuity or corner point on the graph at x = 1?
2. Find all the points on the graph where the function is not differentiable. Determine why the function is not differentiable at these points.
Derivative as a Function Here we construct the graph of the derivative of f(x) from the graph of y = f(x) and vice versa. REMEMBER: When the graph of f (x) has a positive slope, the graph of fx will be above the x-axis. When the graph of f (x) has zero slope the graph of fx will cross the x-axis. When the graph of f (x) has a negative slope, the graph of f x will be below the x-axis.
Sketch the graph of y = f (x) given the graph of y = f(x) 1. f(x) = x 2 y = f (x)
Sketch the graph of y = f (x) given the graph of y = f(x) 2. y = f(x) y = f (x)
Note: For polynomials the degree of the derivative is always one less (one less bump) than the degree of the original polynomial
Sketch the graph of y = f (x) given the graph of y = f(x) 3. y = f(x) y = f (x)
Sketch the graph of y = f (x) given the graph of y = f(x) 4. y = f(x) y = f (x)
Sketch the graph of y = f (x) given the graph of y = f(x) 5. y = f(x) y = f (x)
Sketch the graph of y = f (x) given the graph of y y = f(x) 6. x
Practice
Text Page 92 #3, 4, 9, 11
Construct the graph of f(x) given fx y f ( 1) 2 x
2. Construct the graph of f(x) given fx y f (1) 3 x
3. Construct the graph of f(x) given fx y x
Part B C4. Apply derivative rules including: Constant Rule Constant Multiple Rule Sum Rule Difference Rule Product Rule Quotient Rule Power Rule Chain Rule
Details We will derive the Constant Rule, Sum and Difference Rules, Product and Quotient Rules but these proofs are not required for assessment. We will determine derivatives of functions, using the Constant, Constant Multiple, Power, Sum, Difference, Product and Quotient Rules.
We will determine second and higher-order derivatives of functions. We will determine derivatives of functions using the Chain Rule. We will solve problems involving derivatives drawn from a variety of applications, limited to tangent and normal lines, straight line motion and rates of change.
Computation of derivatives.
Rules for Differentiation Constant Rule: If f(x) = c, where c is a constant, then f( x) 0 or d c 0 dx Graphically: Consider y = 2 What is the slope of the tangent? 0
Proof: f ( x h) f ( x ) f( x) lim h0 h c c 0 lim lim h0 h h 0 h lim 0 0 h0 Example: Find the derivative of: A) f(x) = 5 B) y = -3 y C) D) y = 0
General Power Rule If n is any REAL number then d x dx n nx n1
Examples: 1. 5 A) If f ( x ) x then f( x) 100 B) If y x then y
Examples: 6 C) If y t then dy dt D) d r 3 dr
Examples: 2. Differentiate: A) 1 f( x) 3 x We must rewrite the function as a power of x B) f ( x ) x
Examples: 2. Differentiate: C) f ( x ) 3 x 2 D) f ( x ) x
Note: The Power Rule enables us to find slopes of tangent lines without having to resort to the definition of a derivative.
Constant Rule (II) (Or the Constant Multiplier Rule) If g(x) = c f(x) then g x where c is a constant PROOF: ( ) cf ( x ) g( x h) g( x ) g( x) lim h0 h cf ( x h) cf ( x ) g( x) lim h0 h f ( x h) f ( x ) c lim h0 h cf ( x )
In Leibniz notation: d dx d dx cf x c f x
Example: Differentiate: 3 A) f ( x ) 2x B) f ( x ) 4 x 3 C) f( x) 12x 3 1 4
Text Page 105 # 1, 2, 4, 11, 12, 18
Sum and Difference Rule The next rule tells us that the derivative of a sum (or difference) of functions is the sum (or difference) of the derivatives. If f(x) and g(x) are both differentiable functions then so is f ( x ) g( x ) AND f ( x ) g( x ) f ( x ) g ( x ) OR d f ( x ) g( x ) d f ( x ) d g( x ) dx dx dx
Proof: (of difference rule) Let k(x) = f(x) g(x) k ( x h) k ( x ) k( x) lim h0 h f ( x h) g( x h) ( f ( x ) g( x )) k( x) lim h0 h f ( x h) g( x h) f ( x ) g( x ) lim h0 h f ( x h) f ( x ) g( x h) g( x ) lim h0 h
f ( x h) f ( x ) lim h lim h0 h0 g( x h) g( x ) h f ( x ) g( x )
Note: This rule can be extended to any number of functions f ( x ) g( x ) k ( x ) s ( x ) f ( x ) g( x ) k ( x ) s( x )
Examples 1.
d dx 2. x 2 ( x 1)
d dx 3. 2x 3 3
4. dx 4 d x x x 2
5. dx 3 2 d x x x 4 x
Text Page 105 # 3, 5, 6, 10, 16, 19, 26
Product Rule The following formula was discovered by Leibniz and is called the Product Rule. If P(x) = f(x)g(x) and both f(x) and g(x) are differentiable then: d f ( x ) g( x ) d f x g x f x d g x dx dx dx OR ( ) ( ) ( ) ( ) ( ) P x f x g x f x g x
In words, the Product Rule says that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
PROOF: P ( x h) P ( x ) P( x) lim h0 h f ( x h) g( x h) f ( x ) g( x ) P( x) lim h0 h Math Magic Introduce : f ( x h) g( x ) OR g( x h) f ( x ) f ( x h) g( x h) f ( x h) g( x ) f ( x h) g( x ) f ( x ) g( x ) lim h0 h f ( x h)[ g( x h) g( x )] g( x )[ f ( x h) f ( x )] lim h0 h
f ( x h)[ g( x h) g( x )] g( x )[ f ( x h) f ( x )] lim lim h h h0 h0 [ g( x h) g( x )] [ f ( x h) f ( x )] lim f ( x h)lim g( x )lim h0 h0 h h0 h P ( x ) f ( x ) g( x ) g( x ) f ( x ) OR P ( x ) f ( x ) g( x ) f ( x ) g( x )
Examples Differentiate: A) P(x) = (3x + 2)(4x 3) OLD WAY Expand P(X) = 12x 2 x 6 P ( x ) 24x 1 Product Rule: P ( x ) (3x 2) (4x 3) (3x 2)(4x 3) 3(4x 3) (3x 2)4 12x 9 12x 8 24x 1
NOTE: The product rule is more useful for finding the derivative of functions like: y = x 2 sinx y x x 1 x y e cosx However we need to have the transcendental rules first!
Examples: Differentiate B) 2 y x x 2
C) Suppose f(2) = 5, f (2) = 6, g(2) = 7 and g (2) = -1, find (fg) (2) ( fg) (2) f (2) g(2) f (2) g(2)
D) If h(x) = xg(x) and it is known that g(3) = 5 and g(3) = 2, find h(3).
Quotient Rule If Qx ( ) f ( x ) gx ( ) and both f(x) and g(x) are differentiable then: f ( x ) g( x ) f ( x ) g ( x ) Q ( x) 2 gx ( )
In words, the Quotient Rule says that the derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
PROOF: Use the product rule! Qx ( ) f( x) f ( x ) Q( x ) g( x ) gx ( ) ( ) ( ) ( ) ( ) ( ) f x Q x g x Q x g x Solve for Q( x) ( ) ( ) ( ) ( ) ( ) Q x g x f x Q x g x f( x) Q( x ) g( x ) f ( x ) g( x ) gx ( )
gx ( ) f( x) Q ( x ) g( x ) f ( x ) g ( x ) g( x ) g( x ) Q( x ) g( x ) Q( x) f ( x ) g( x ) f ( x ) g( x ) gx ( ) f ( x ) g( x ) f ( x ) g( x ) gx ( ) 2 This proof can also be done using the definition of derivative in the same way that the product rule was proven!
Examples: Differentiate. x 2 A) y x 1
B) y 3x 1 2x 5
C) y 2x x 2 2 x 1
D) y 6 1 x x
Higher Derivatives
Higher Derivatives If f is a differentiable function, then its derivative f is also a function, so f may have a derivative of its own, denoted by (f) = f. This new function f is called the second derivative of f because it is the derivative of the derivative of f.
The Second Derivative d f ( x ) f ( x ) dx f ( x h) f ( x ) f( x) lim h0 h : Using Leibniz notation, we write the second derivative of as y = f (x) as : d d f ( x ) f ( x ) dx dx 2 d ( ) 2 f x dx 2 d y dx 2
Find the 2 nd derivative of: 1. f(x) = x 2 + 2x
2. f(x) = x 3 3x
3. s(t) = -4.9t 2 + 50 t + 200
Higher Derivatives The 2 nd derivative has applications in curve sketching and physics. In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration. The instantaneous rate of change of velocity with respect to time is called the acceleration of the object.
Example A robot moves along a straight line according to the function: s t t t t 3 2 ( ) 5 8 10 A) Determine when the robot is stopped. B) Determine the acceleration when the robot is stopped
Practice: Text Page 105 #43
In curve sketching the 2 nd derivative tells you how fast the slopes of the tangents are changing. This is called concavity.
Example: 1. y = x 2 Find y y = 2x y = 2 The fact that the 2 nd derivative is positive tells you that the tangent lines are increasing. It also tells you that the graph is concave upwards. Holds water Or that all of the tangent lines lie below the curve
Example: 2. y = -x 2 3x +4 Find y y = -2x - 3 y = -2 The fact that the 2 nd derivative is negative tells you that the tangent lines are decreasing. It also tells you that the graph is concave downwards. Does NOT Hold water Or that all of the tangent lines lie above the curve
Sketch the graph of y = f (x) and y = f (x) given the graph of y = f(x) y = f(x) y = f (x) y = f (x)
Sketch the graph of y = f (x) and y = f (x) if f(x) = sin x y = f(x) y = f (x) y = f (x) y y y x x x
Higher Derivatives The third derivative f is the derivative of the second derivative f ( x ) f x So y f ( x ) can be interpreted as the slope of the curve y = f (x) or as the rate of change of f (x).
Higher Derivatives If y = f(x), then alternative notations for the third derivative are d 2 d y d 3 y 2 dx dx dx 3 The process can be continued but the multiple prime notation will eventually become cumbersome for higher derivatives. y f x The fourth derivative f is usually denoted by f (4).
In general, the nth derivative of f is denoted f (n) by and is obtained from f by differentiating n times. If y = f(x), we write n y n f x n d y dx n
Note: The superscripts used for higherorder derivatives, such as y (4) for the fourth derivative of y, are not exponents. The presence of the parenthesis in the exponent denotes differentiation 2 (i.e., f x f x ) while the absence of the parenthesis denotes exponentiation (i.e., f 2 x f x 2 ).
Examples 1.Determine the first five derivatives 4 3 2 of f ( x ) x 2x 2x 3x 4
7 5 8 2.Given f ( x ) 3x 2x 7x 9 determine the value of n, such that n 0 f x
3.Determine the point on the graph of 3 2 y 2x 12x 5x 6 where 2 d y 0 2 dx
DERIVATIVES OF COMPOSITE FUNCTIONS CHAIN RULE IMPLICIT DIFFERENTATION
Chain Rule Differentiate: A) (x + 1) 2
B) (x + 2)3
C) (x2 + 1)3
Lets break C) into a composition of functions f(g(x)) What would be the outer function f(x)? What would be the inner function g(x)? Therefore f(g(x)) = f(x 2 + 1)= (x 2 + 1) 3 And f ( x ) Which means x 3 g( x ) x 2 1 d f ( g ( x )) dx 3x 2 1 2 2x d f ( g ( x )) f ( g ( x )) g ( x ) dx
Basically the Chain Rule is a way of finding the derivative of a composition of functions Chain Rule: If C ( x ) f g( x ) f ( g( x )) C ( x ) f ( g( x )) g( x ) then derivative of the outer function wrt the inner function derivative of the inner function
Lets consider derivatives as IROC Example: A car travels twice as fast as a bicycle A bicycle travels 4 times as fast as a person walking A car travels 8 times as fast as a person walking Numerical example: A person walks at 5 km/h The bicycle travels at 20 km/h The car travels at 40 km/h 8 times as fast
This can be applied to any situation involving related IROCS dy du If y changes a times as fast as u Outer function du dx Inner function and u changes b times as fast as x then y changes ab times as fast as x dy dx ab dy dy du dx du dx Leibniz notation for the Chain Rule
Referencing back to C) y = (x 2 + 1) 3 how can the Leibniz notation for the Chain rule be applied to this problem? What is the inner function for y? x 2 +1 The function u is the inner function Rewrite y = (x 2 + 1) 3 using u y = (u) 3 Differentiate using dy dy dx dy du du dx d du dy du dx du dx d dx 3 2 u x 1 2 3u 2x 2 x 3 1 2x
Examples: 1.Find A) d dx x 1 2 B) d dx 2x 1 3 C) d dx x 2 1 50
D) d dx x 2 x E) d dx x 2x 5 4
2.Find A) d 2 dx x x 3 2 4 1 2
d dx 3 B) 4x 4x 1 5 4 2
3.A) If y u u 3 3 2 5, where u x 3 find dy dx
3.B) If y = u 10 7u 3 + 1, where u = 3x 2 2x, find dy dx
4. If h(x) = f(g(x)), find h(2) when g(2) = 3, g(2) 5, f (2) 1, f (3) 2 and f (5) 4
Text Page 120-122 # 1, 2, 7, 8, 9, 11, 53, 54,55, 56
Fun Examples Differentiate the following functions: 5 A) f ( x ) 2x 1 4x 1
B) y 2x 3 3 4x 7
3 C ) f ( x ) 2x 3x
D) f ( t ) 2 t 1 t 1 10
dy 2. Determine the value of where dy 5, dv 2 x and du 3 du dx dv dx
Solve problems involving derivatives drawn from a variety of applications limited to tangent and normal lines, straight line motion and rates of change.
Tangent and Normal Lines 1. Find the equation of the tangent 1 line to the curve y at the 4 1 20 x point 2, 2
2. Determine the equation of the tangent line to the curve at the given value. 3 2 A) y x x x 1, at x 1
3 2 B) y x 2, at x 2
C) y 7 x 3 x, at x 1
3. Find the equation of the normal line to the curve 3 s( t ) t 2t 5, at x 11
4. At what point on the curve 4 y x x 25 2 is the tangent parallel to the line 7x y 2?
5. Find the equation of both tangent lines that pass through the origin and are tangent to the parabola y = x 2 + 1 (0, 0) (x, x 2 + 1) How do we find the slope of the tangent line? y2 y1 mtan x x 2 1 2 x 10 f( x) x 0 2 x 1 2x x 2 2 2x x 1 2 x 1 0 x 1
Thus the slopes of the tangents are: For x = -1 m tan = 2x = 2(-1) = -2 For x = 1 m tan = 2x = 2(1) = 2 And the equation is: (y 0) = -2(x 0) y = -2x And the equation is: (y 0) = 2(x 0) y = 2x
2 y x x 6. (YOUR TURN) Find the equations of the 2 tangent lines to the parabola y x x that pass through the point 2, 3. Sketch the curve and the tangents
y x
7. Find the x-coordinates of the points on the hyperbola xy 1 where the tangents from the point 1, 1 intersect the curve.
Text Page 120-122 #17, 21, 26, 48, 75
Straight Line Motion velocity (rate of change of displacement with respect to time) If an object travels along a path, and displacement s(t) is measured over time, then the first derivative s'(t), equals the velocity v(t) of that object. (rate acceleration of change of velocity with respect to time) The second derivative s"(t) = v'(t) equals the acceleration a(t) of that object.
For Straight Line Motion Problems, when v(t) = 0 object is not moving v(t) > 0 (positive) object is moving forward (to the right) v(t) < 0 (negative) object is moving backward (to the left)
For Straight Line Motion Problems, when no change in the velocity a(t) = 0 there is a(t) > 0 (positive) object s velocity is increasing a(t) < 0 (negative) object s velocity is decreasing
NOTE: If the signs of v(t) and a(t) are the same (both positive or both negative),then the speed of the object is increasing If the signs of v(t) and a(t) are opposite (one positive and the other negative), then the speed of the object is decreasing
Example A particle moves back and forth along a horizontal line defined by the position function: s(t) = t 3-12t 2 + 36t - 30, t 0. (a) Determine the velocity and acceleration functions. 2 v( t ) s t 3t 24t 36 a( t ) v t 6t 24 (b) When is the velocity zero? When is the acceleration zero? vt ( ) 0 at ( ) 0 2 3t 24t 36 0 2 3( t 8t 12) 0 3 t 2 t 6 0 t 2or t 6 6t 24 0 t 4
(c) During what time intervals is the velocity positive? Explain what it means. vt ( ) 0 t t 3 2 6 0 [0,2) (6, ) The particle is travelling to the right from t = 0 seconds to t = 2 seconds and after 6 seconds (d) During what time interval is the velocity negative? Explain what it means. vt ( ) 0 t t 2,6 3 2 6 0 The particle is travelling to the left from t = 2 seconds to t = 6 seconds
(e) When is the particle speeding up? Slowing down? This depends on the acceleration of the particle and the direction it is travelling. at ( ) 0 6t 24 0 6t 24 and at ( ) 0 t 4 t 4 (f ) Provide an overall description of what is happening to the particle. So for the first 2 seconds, the particle is travelling right but slowing down. At 2 seconds it stops and then speeds up in as it travels left. At 4 seconds the acceleration is zero and the particle has reached its max speed in the left direction.
After 4 seconds the acceleration is positive and thus the particle slows down and stops travelling left at 6 seconds. After 6 seconds the particle travels right and speeds up.
(g) What is the total distance travelled by the particle during the first minute?
Text Page 105-6 #43, 44 a) b), 50 Page 142 # 65
Warm UP A demographer develops the function P(x) = 12500 + 320x - 0.25x 3 to represent the population of the town Tanville x years from now. (a) Determine the population of the town today. (b) Predict the instantaneous rate of change of the population during year 3.
Warm UP P(x) = 12500 + 320x - 0.25x 3 (c) Determine the average rate of change in the population between years 2 and 7. (d) In what year can you expect the population to increase by 245 people? (e) Is the rate of change in the population increasing or decreasing? Explain.
Part C C5. Determine the derivative of a relation, using implicit differentiation. Determine the derivative of an implicit relation. Determine the equation of the tangent and normal line to the graph of a relation at a given point. Determine the second derivative of a relation, using implicit differentiation.
Implicit Differentiation Function y = 2 y = c y = x y = x 2 y = (5x+1) 2 y = u 2 Explicitly Defined Derivative dy 0 dx dy 0 dx dy 1 dx dy 2x dx y y 2 5x 1 5 2u u x = y 2 This is an implicitly defined relation. 1 2yy
NOTE: x = y 2 can be solved explicitly and then differentiated.
One of the first implicit relations you encountered was the equation of a circle: x 2 + y 2 = r 2 Suppose you are asked to find the slope of the tangent line to the circle centred at the origin with a radius of 10 at the point (6, 8) You would need to differentiate the equation of the circle and substitute x = 6 into the derivative. What is the equation for this circle?
Lets find dy dx for x 2 + y 2 = 10 2 both explicitly and implicitly y 100 x 2 x y 2 2 100 dy dx 2 dy x dx 2 100 x 1 1 100 2 2 x 2 x dy 2x 2y 0 dx dy 2y 2x dx dy 2x x dx 2y y Sub in the point (6, 8) dy dx 6 100 6 dy 6 6 3 dx 64 8 4 2 dy dx 6 3 8 4
Note: The derivative expression for explicit differentiation involves x only, while the derivative expression for implicit differentiation may involve both x and y. If you have a choice you can solve by implicit differentiation or by first solving the given equation for y, however you are encouraged to choose the most efficient method.
Note: There are, however, examples where it is impossible to solve the equation for y as an explicit function of x. In these cases, the method of implicit differentiation becomes the only choice for solving for y'.
Examples: 1. Find y A) xy = 1
B) x 3 + 2x + y + y 3 = 3
C) x 2 + xy + y 2 = 5
2. Find for 2x 3 u x u 2 = u x 2 du dx
3. Find the equation of the A) tangent and B) normal lines to the curve 3x 2-2xy + xy 3 = 7 at the point (1, 2)
4. Find y A) x 2 y 2 y 3 = xy
B) x y xy 6
5. PROVE: when x 3 + y 3 = 10 y 20x 5 y
6. Prove that the tangent to a circle at a point P is perpendicular to the radius drawn to P m m (a,b) C r P (x,y) Show y mcp x tan y x 2 1 2 1 CP y b x a The perpendicular slope is: m CP This is what should dx equal dy x a y b Equation of circle is: (x a) 2 + (y - b) 2 = r 2 dy Find dx
7. The curve 2(x 2 + y 2 ) 2 = 25(x 2 - y 2 ) shown below is called a lemniscate. Determine the equation of the tangent line to this curve at (3,1).