# Differentiation. The Derivative and the Tangent Line Problem.

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Differentiation

The Derivative and the Tangent Line Problem

The Tangent Line Problem To find the tangent line of a curve at a point P, you need to find the slope of the tangent line at point P. You can approximate this slope using the secant line through the point of tangency and a second point on the curve where (c, f(c)) is the point of tangency and (c + ∆x, f (c + ∆x)) is a second point on the graph of f. Slope formula: m = (y 2 –y 1 )/(x 2 -x 1 ) = ∆y/ ∆x Slope formula: m sec =( f (c + ∆x) – f(c))/(c + ∆x – c) m sec = f (c + ∆x) – f(c))/ ∆x

Definition of Tangent Line with Slope m If f is defined on an open interval containing c, and if the limit lim ∆y/ ∆x = lim (f(c + ∆x) – f(c))/ ∆x = m ∆x  0 Exists, then the line passing through (c,f(c)) with slope m is the tangent line to the graph of f at the point (c,f(c)) The slope of the tangent line is also called the slope of the graph of f at x =c

Example Find the slope of the graph of f(x) = 2x -3 at the point (2,1) Use the definition of Tangent Line with Slope m Then c = 2 lim ∆y/ ∆x = lim (f(c + ∆x) – f(c))/ ∆x = m ∆x  0 lim (f(2 + ∆x) – f(2))/ ∆x = {[2(2 + ∆x) -3]-[2(2) -3]}/ ∆x ∆x  0 = lim (4 + 2 ∆x - 3 – 4 + 3)/ ∆x = 2 ∆x / ∆x = 2 ∆x  0 m = 2

Example Find the slope of the tangent line to the graph of 1. g(x) = 3x – 6 at the point (2,0) 2. g(x) = -4x + 1 at the point (-1,5) 3. f(x) = x 2 + 1 at the point (-1,2) 4. f(x) = x 2 + 1 at the point (0,1)

Definition of the Derivative of a Function The derivative of f at x is given by f′(x)= lim (f(x + ∆x) – f(x))/ ∆x ∆x  0 Provided the limit exists. For all x for which this limit exists, f′ is a function of x. f′ is a new function of x which gives the slope of the tangent line to the graph of f at the point (x,f(x))

Differentiation Differentiation the process of finding the derivative of a function. A function is differentiable at x if its derivative exists at x and is differentiable on an open interval (a,b) if it is differentiable at every point in the interval

Notations for Derivatives f′(x) dy/dx y′ d/dx[f(x)] D x [y]

Derivatives dy/dx = lim ∆y/ ∆x ∆x  0 = lim (f(x + ∆x) – f(x))/ ∆x ∆x  0 = f′(x)

Example Find the derivative by the limit process 1. g(x) = 2x 2 - 9 2. f(x) = x 3 + 2x 3. f(x) = √x 4. y = 2/t

Differentiability and Continuity The following alternative limit form of the derivative is useful in investigating the relationship between differentiability and continuity. The derivative of f at c is f′(c) = lim (f(x) –f(c))/(x – c) x  c provided the limit exists. The existence of the limit in this alternative form requires that the one-sided limits exist and are equal

Differentiability and Continuity – con’d lim (f(x) –f(c))/(x – c) x  c- and lim (f(x) –f(c))/(x – c) x  c+ These one-sided limits are called the derivatives from the left and from the right. It follows that f is differentiable on the closed interval [a,b] if it is differentiable on (a,b) and if the derivative from the right at a and the derivative from the left at b both exist.

Differentiability and Continuity – con’d Differentiability implies continuity. If a function is differentiable at a point, then the function is continuous at that point Continuity does not imply differentiability.  A function may be continuous at a point, but not differentiable at that point  A function that is not continuous at a point is not differentiable at that point

Example Determine differentiability and/or continuity 1. g(x) = |x – 2| at x = 2 2. f(x) = x 1/3 at x = 0 3. f(x) = [| x |] at x = 0 the greatest integer

Basic Differentiation Rules and Rates of Change

The Constant Rule The derivative of a constant function is 0. That is, if c is a real number, then d/dx[c] = 0

Example Find the derivative of these functions 1. y = 7 2. f(x) = 0 3. s(t) = -3 4. y = kπ 2, where k is constant

The Power Rule If n is a rational number, then the function given by f(x) = x n is differentiable and d/dx[x n ] = nx n-1 For f to be differentiable at x=0, n must be a number such that x n-1 is defined on an interval containing 0.

The Power Rule when n = 1 If n =1, then the function given by f(x) = x is differentiable and d/dx[x] = 1

Example Find the derivatives of these functions 1. f(x) = x 3 2. f(x) = 3 √x 3. s(t) = 1/x 2 4. y = x 4

Finding the Equation of a Tangent Line  First, evaluate the function at x  Second, find the slope using the derivative and evaluate at x  Third, use point slope form to find the tangent line

Example Find the equation of a tangent line for each of these functions 1. f(x) = x 2 when x = -2 2. f(x) = 3 √x when x = 8 3. s(t) = 1/x 2 when x = -3 4. y = x 4 when x = ½

The Constant Multiple Rule If f is a differentiable function and c is a real number, then cf is also differentiable and d/dx[cf(x)] = cf′(x)

Example Find the derivatives of these functions 1. f(x) =2/x 2. f(x) =4t 2 /5 3. s(t) =2√x 4. y = -3x/2 5. 1/2( 3 √x 2 ) 6. y = 7/(3x) -2

The Sum and Difference Rules The sum (or difference) of two differentiable functions f and g is itself differentiable. Moreover, the derivative of f + g (or f - g) is the sum (or difference) of the derivatives of f and g. d/dx [f(x) + g(x)] = f′ (x)+ g′ (x) Sum Rule d/dx [f(x) - g(x)] = f′ (x) - g′ (x) Difference Rule

Example Find the derivatives of these functions 1. f(x) =x 3 – 4x + 5 2. f(x) =3x 2 - 4 3. s(t) =-2x 3 + 9x 2 - 2 4. y = -x 4 /2 + 3x 3 – 2x

Rates of Change The derivative can be used to determine the rate of change of one variable with respect to another. A common use of rate of change is to describe the motion of an object moving in a straight line. The function s that gives the position (relative to the origin) of an object as a function of time t is called a position function. Then rate = distance/time or

Rates of Change – cont’d Average velocity is Change in distance/change in time = ∆s/ ∆t which is the derivative

Example – Finding Average Velocity If a billiard ball is dropped from a height of 100 feet, its height s at time t is given by the position function s = -16t 2 +100 Find the average velocity over each time interval 1. [1,2] 2. [1,1.5] 3. [1,1.1] What do the negative answers indicate?

Example – Using Derivative to Find Velocity At time t = 0, a diver jumps from a platform diving board that is 32 feet above the water. The position of the diver is given by s(t) = -16t 2 + 16t + 32 where s is measured in feet and t is measured in seconds a) When does the diver hit the water? b) What is the diver’s velocity at impact?

The Product and Quotient Rules and Higher-Order Derivatives

The Product Rule The product of two differentiable functions f and g is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first. d/dx [f(x)g(x)] = f(x)g′(x) + g(x)f′(x)

Example Find the derivative of each function 1. h(x) =(3x-2x 2 )(5 + 4x) 2. y = (3x -1)(2x + 5) 3. g(x) = (x 3 +2x 2 -5)(x 4 – 4x 2 ) 4. f(x) = x 2 (x+1)(2x-3) 5. y= (2x 3 + 5x)(7)

The Quotient Rule The product of two differentiable functions f and g is itself differentiable at all values of x for which g(x) ≠ 0. Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. d/dx [f(x)/g(x)] = g(x)f′(x) - f(x)g′(x)] ÷ [g(x)] 2

Example Find the derivative of each function 1. h(x) = (5x -2) ÷ (x 2 + 1) 2. y = (x 2 + 3x) ÷ 6 3. g(x) = (12x 3 - 3x 2 ) ÷ (6x 2) 4. f(x) = (x+1) ÷ (x-1) 5. y = (x-3) ÷ (x 2 -4)

Example – Find an Equation of the Tangent Line f(x) = (3 – 1/x) ÷ (x+5) at (-1,1) 1. Find f′(x) 2. Evaluate f′(x) 3. Use point-slope form to find equation

Higher-Order Derivatives s(t) Position function v(t) = s′(t) Velocity Function a(t) = v′(t)= s″(t) Acceleration Function

Example – Finding Acceleration A car’s velocity starting from rest is v(t) = 100t ÷ (2t+15) where v is measured in feet per second. Find the acceleration at each of the following times. 1. 5 seconds 2. 10 seconds 3. 20 seconds

The Chain Rule

The chain rule is a very powerful differentiation rule that deals with composite functions

The Chain Rule Without the Chain RuleWith the Chain Rule  f(x) =x 2 + 1  g(x)=3x + 2  y =x + 2  y = (x 2 +1) ½  f(x) = 3x +2) 5  g(x) = (x + 2) 1/3

The Chain Rule If y =f(u) is a differentiable function of u and u =g(x) is a differentiable function of x, then y =f(g(x)) is a differentiable function of x and dy/dx = dy/du · du/dx or d/dx[f(g(x))] = f′(g(x))g′(x)

Example – Applying the Chain Rule  Write the function as a composite of functions  Apply Chain Rule y = f(g(x))u = g(x)y =f(u) y= 1/(x+1)u = x + 11/u y = (3x 2 – x +1) ½ u = 3x 2 – x +1u ½

Example – Applying the Chain Rule  Find dy/dx for y = (x 2 +1) 3 y = f(g(x))u = g(x)y =f(u) y = (x 2 +1) 3 u = x 2 +1u 3 dy/dx = dy/du · du/dx 3(x 2 + 1) 2 (2x) = 6x(x 2 +1) 2 dy/du du/dx

The General Power Rule If y = [u(x)] n where u is a differentiable function of x and n is a rational number, then dy/dx = n[u(x)] n-1 du/dx or d/dx[u n ] = nu n-1 u′

Example Differentiate the functions 1. f(x) = (3x -2x 2 ) 3 2. f(x) = [(x 2 -1) 2 ] 1/3 3. g(t) = -7/(2t-3) 2 4. y = [(3x -1)÷ (x 2 +3)] 2

Implication Differentiation

Implicit and Explicit Functions Up to this point, most in explicit form. For example in the equation y = 3x 2 - 5 the variable y is explicitly written as a function of x Implicit FormExplicit FormDerivative xy = 1y = 1/x-1/x 2 Sometimes, this procedure doesn’t work, so you would use implicit differentiation

Implicit Differentiation To understand how to find dy/dx implicitly, you must realize that the differentiation is taking place with respect to x. This means when you differentiate terms involving x alone, you can differentiate as usual, but when you differentiate terms involving y, you must apply the Chain Rule, because you are assuming that y is defined implicitly as a differentiable function of x.

Differentiating with Respect to x 1. d/dx[x 3 ] = 3x 2 2. d/dx[y 3 ] = 3y 2 dy/dx where y =u and n = 3 nu n-1 u′ 3. d/dx[x +3y] = 1 +3(dy/dx) 4. d/dx[xy 2 ]= xd/dx +y 2 (d/dx)[x] =2xy(dy/dx) + y 2

Guidelines for Implicit Differentiation 1. Differentiate both sides of the equation with respect to x. 2. Collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation. 3. Factor dy/dx out of the left side of the equation. 4. Solve for dy/dx

Example Find dy/dx given that y 3 +y 2 -5y –x 2 = - 4

Example – Finding the Slope of a Graph Implicitly Find the slope of the tangent line to the graph of 3(x 2 + y 2 ) 2 = 100xy at the point (3,1) Find dy/dx and then use (3,1) to find the value of the slope

Other Applications of Implicit Differentiation 1. Finding a Differentiable Function 2. Finding the Second Derivative Implicitly 3. Finding a Tangent Line to a Graph

Related Rates

Guidelines for Solving Related-Rate Problems 1. Identify all given quantities and quantities to be determined. Make a sketch and label the quantities. 2. Write an equation involving the variables whose rates of change either are given or are to be determined. 3. Using the Chain Rule, implicitly differentiate both sides of the equation with respect to time t. 4. Substitute into the resulting equation all known values for the variables and their rates of change. Then solve for the required rate of change.

Example An airplane flying at an altitude of 6 miles is on a flight path that will take it directly over a radar tracking station. If distance s is decreasing at a rate of 400 miles per hour when s = 10 miles, what is the speed of the plane? Draw and label a right triangle. s is the hypotenuse of the right triangle and x is the horizontal distance from the station.

Example – Cont’d s = 10 x = (10 2 – 6 2 ) ½ = 8 ds/dt = -400 when s =10 Find dx/dt when x=10 and x = 8 The function is x 2 + 6 2 =s 2 1. Differentiate with respect to x 2. Solve for dx/dt 3. Substitute for s,x and ds/dt

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