AP Calculus AB Class Notes 3.7 Implicit Differentiation pp 157 - 164 Objectives: Take the derivative implicitly. Use all previous rules (sum & difference rule, constant multiple rule, power rule, product rule, quotient rule, chain rule, derivatives of trig functions) in your practice of implicit differentiation. Outline Determine the derivative of a relation by differentiating implicitly. Find the equations of lines tangent to and normal to an implicit curve. Find higher order derivatives using implicit differentiation. Vocabulary implicit differentiation normal to a surface Power Rule for Rational Powers of x Omissions none Day 1 Introduction to Implicit Differentiation Implicit differentiation is a trick that allows us to differentiate relations. As you may recall, when we have graph that violates the vertical line test, it is not a function, but rather it is a relation. An example of a relation is ๐ฅ 3 + ๐ฆ 3 − 9๐ฅ๐ฆ = 0 The graph of this relation looks like the following: y As you can see, this relation๏ธ clearly violates the vertical line test; it is not a function. Furthermore, it cannot easily be broken down to two functions. Nonetheless, we can see that the relation has a tangent everywhere except where the tangent is ๏ทhorizontal. ๏ถ Implicit differentiation allows us to find the derivative without putting the relation into function form. ๏ต ๏ด How do we differentiate a relation that is dependent on x and y? It’s kind of a trick. We apply the chain rule to a function of y. For example: ๏ณ ๏ฒ ๏ฑ x ๏ญ๏ถ ๏ญ๏ต ๏ญ๏ด ๏ญ๏ณ ๏ญ๏ฒ ๏ญ๏ฑ ๏ฑ ๏ญ๏ฑ ๏ฒ ๏ณ ๏ด ๏ต ๏ถ ๐ ๐๐ฆ ๐ฆ= ๐๐ฅ ๐๐ฅ ๏ท I know that that sounds simplistic. It is. ๏ญ๏ฒ ๏ญ๏ณ Let’s try the simple differentiation: ๐ 3 ๐ฆ ๐๐ฅ ๏ญ๏ด To perform this, we treat y as a function, and use the chain rule: we differentiate the outside (the cube), leave the inside alone, and multiply by the derivative of the inside (y). The derivative of the function ๐ฆ 3 is ๐ 3 ๐๐ฆ ๐ฆ ๐๐๐๐๐๐๐ 3๐ฆ 2 ๐๐ฅ ๐๐ฅ Class Notes 3.4 Page 1 of 19 Let’s practice taking the derivative of several different terms: Term ๐ ๐ ๐ y ๐ฆ2 √๐ฆ cos ๐ฆ csc ๐ฆ ๐ฅ๐ฆ ๐ฅ2๐ฆ ๐ฅ๐ฆ 2 ๐ฅ3๐ฆ2 ๐ฅ ๐ฆ ๐ฆ ๐ฅ Class Notes 3.4 Page 2 of 19 ๏ท ๏ถ Find ๐๐ฆ ๐๐ฅ for the relation ๐ฅ 3 + ๐ฆ 3 − 9๐ฅ๐ฆ = 0 ๏ต ๏ด ๏ณ ๏ฒ ๏ฑ x ๏ญ๏ถ ๏ญ๏ต ๏ญ๏ด ๏ญ๏ณ ๏ญ๏ฒ ๏ญ๏ฑ ๏ฑ ๏ฒ ๏ณ ๏ด ๏ต ๏ญ๏ฑ ๏ญ๏ฒ ๏ญ๏ณ ๏ญ๏ด End of 3.7 Day 1 3.7 Day 2 Finding the Second Derivative Implicitly To find the second derivative of a function implicitly, you find the first derivative, and then the second. When you take the second derivative, you’ll have some terms expressed as ๐ฆ′. Substitute-in the first derivative at these places. ๐2 ๐ฆ Example: Find ๐๐ฅ 2 of the following: 3๐ฅ 2 − 4๐ฆ 2 = 7 ๐ [3๐ฅ 2 − 4๐ฆ 2 = 7] ๐๐ฅ 6๐ฅ − 8๐ฆ๐ฆ ′ = 0 −8๐ฆ๐ฆ ′ = −6๐ฅ 3๐ฅ ๐ฆ′ = 4๐ฆ ๐ Now, take the ๐๐ฅ of each side again: ๐ ′ 3๐ฅ [๐ฆ = ] ๐๐ฅ 4๐ฆ 3 ๐ฆ โ 1 − ๐ฅ๐ฆ′ ๐ฆ ′′ = 4 ๐ฆ2 Now, subsititue the first derivative into the second derivative’s equation above. 3๐ฅ ๐ฆ − ๐ฅ [4 ๐ฆ ] 3 ๐ฆ ′′ = 4 ๐ฆ2 Simplify. 3๐ฅ2 3๐ฅ ′′ ๐ฆ = 3 ๐ฆ−๐ฅ[4๐ฆ] 4 ๐ฆ2 = 4๐ฆ2 3๐ฅ2 3 ๐ฆ− 4๐ฆ 3 4๐ฆ − 4๐ฆ =4 ๐ฆ 2 4 ๐ฆ2 2 2 = 3 4 4๐ฆ2 −3๐ฅ2 4๐ฆ ๐ฆ2 3 4๐ฆ 2 −3๐ฅ 2 4๐ฆ 3 =4 Now, recall that the original equation was 3๐ฅ − 4๐ฆ = 7. We can therefore, subsitute this into the numerator and get: 3 4๐ฆ 2 − 3๐ฅ 2 3 −1(3๐ฅ 2 − 4๐ฆ 2 ) 3 −7 −21 = = = 3 3 3 4 4๐ฆ 4 4๐ฆ 4 4๐ฆ 16๐ฆ 3 Class Notes 3.4 Page 3 of 19 ๏ถ ๏ท Let’s try several more: ๐2 ๐ฆ Find ๐๐ฅ 2 ๐ฅ3 + ๐ฆ3 = 1 ๐ฅ3๐ฆ3 − 4 = 0 ๐ฅ๐๐๐ ๐ฆ = ๐ฆ End of 3.7 Day 2 Class Notes 3.4 Page 4 of 19 3.7 Day 3 Applications of Implicit Differentiation The main application of implicit differentiation is finding the tangent and normal lines, and variations on these. Here’s an example: Find the point(s) (if any) of horizontal tangent lines: ๐ฅ 2 + ๐ฅ๐ฆ + ๐ฆ 2 = 6 and write the equation of the tangent lines at thos points. Find ๐ฆ′ if ๐ฅ 3 + ๐ฆ 3 = 6๐ฅ๐ฆ. Find the tangen to the folium of Descartes ๐ฅ 3 + ๐ฆ 3 = 6๐ฅ๐ฆ at the point (3,3). At what point in the first quadrant is the tangen t line horizontal. Class Notes 3.4 Page 5 of 19 Use implicit differentiation to find an equation of the tangent line to the curve at the given point. ๐ ๐ ๐ฆ sin 2๐ฅ = ๐ฅ cos 2๐ฆ, ( , ) 2 4 At what point(s) is the tangent line to the curve ๐ฆ 2 = 2๐ฅ 3 perpendicular to the line 4๐ฅ − 3๐ฆ + 1 = 0 ? End of 3.7 Day 3 Class Notes 3.4 Page 6 of 19 AP Calculus AB Class Notes 3.8 Derivatives of Inverse Functions pp 165 - 171 Objectives: In this section we’ll find the value of the derivative of the inverse of a function without finding the analytical derivative of the inverse function or even knowing the inverse function itself. We’ll set this up graphically, and then memorize the analytical formula. Next, we’ll memorize and apply the derivatives of the three main trig functions: arcsin x, arccos x, and arctan x. Outline Draw the graphs that illustrate the relationship between the slope of the tangent of a function with the slope of its inverse. Use formulae to find the derivative of arcsin, arccos, and arctan. Vocabulary implicit differentiation normal to a surface Power Rule for Rational Powers of x Omissions none 3.8 Day 1 The Derivative of an Inverse Function We learned in earlier classes that we can find the inverse of many functions by reversing the x and y variables and solving for y. But in many cases, you cannot analytically find the inverse function. For example, the function: ๐ฆ = 3๐ฅ 2 − 2๐ฅ + 5 does not have an inverse as a function of x. However, we can use what we know about inverse functions to find the value of the derivative without knowing the inverse function. We’ll start with a property of the behavior of the derivative of the inverse of a function. Graphs of inverses have reciprocal slopes. ๐(๐ฅ) = ๐ฅ 2 ๐(๐ฅ) = ๐ฅ 3 3 ๐ −1 (๐ฅ) = √๐ฅ ๐ −1 (๐ฅ) = √๐ฅ ๐(๐ฅ) = ๐ ๐ฅ ๐ = ln ๐ฅ −1 (๐ฅ) ๐(๐ฅ) = cos ๐ฅ ๐ −1 (๐ฅ) = cos−1 ๐ฅ In the next example, we’ll see how the slopes of inverse functions are reciprocals. Class Notes 3.4 Page 7 of 19 For the following functions and inverses, find the domains of ๐ and ๐ −1 . Find the ranges. Graph each. That the slopes of the graphs of ๐ and ๐ −1 are reciprocals at the given points. 1 1 ๐(๐ฅ) = 3 − 4๐ฅ (2 , 8) ๐ −1 (๐ฅ) = 3−๐ฅ 4 1 1 (8 , 2 ) ๐ domain__________range______________ ๐ −1 domain__________range______________ ๐(๐ฅ) = √๐ฅ − 4 (5,1) ๐ −1 (๐ฅ) = ๐ฅ 2 + 4 (1,5) ๐ domain__________range______________ ๐ −1 domain__________range______________ ๐(5) =________________________ 1 2 ๐ ( ) =________________________ ๐ −1 (1) =________________________ 1 ๐ −1 (8) =________________________ Here’s a closer look at the phenomenon that slopes of inverses are reciprocal. For the sake of this discussion, let’s start with a generic function, ๐(๐ฅ) and it’s inverse, ๐(๐ฅ). Follow the steps. ๏ท Draw the arbitrary ๐(๐ฅ). ๏ท Reflect ๐(๐ฅ) in the line ๐ฆ = ๐ฅ to find its inverse. ๏ท Label a point on ๐ with the point (๐ฅ, ๐(๐ฅ)). ๏ท Draw the tangent to f through (๐ฅ, ๐(๐ฅ)). The slope of this line is m. ๏ท Now, find the inverse reflection, (๐(๐ฅ), ๐ฅ) on the function g. ๏ท Draw the tangent to g through (๐(๐ฅ), ๐ฅ). The slope of this 1 line is ๐ ๏ท Since the two derivatives are inverses, we can say that the 1 derivative of ๐(๐ฅ) = ′ . This you must memorize. ๐ (๐(๐ฅ)) Class Notes 3.4 Page 8 of 19 Let’s do the same thing again, this time with g being our the function g. Type equation here. ๐ ′ (๐ฅ) = ๐′ (๐ฅ) = 1 ๐ ′ (๐(๐ฅ)) 1 ๐′ (๐(๐ฅ)) Memorize This! Now, let’s use it. Recall the function ๐(๐ฅ) = 3๐ฅ 2 − 2๐ฅ + 5 −1 Use this function to find (๐ )′(6). Throughout this unit, we’ll label ๐ −1 (๐ฅ) as ๐(๐ฅ). Find ๐′(6). From the formula, ๐(๐ฅ) = 1 . ๐ ′ (๐(6)) First of all, what is ๐(6)? Since ๐(1) = 6, ๐(6) = 1. Since we know what ๐ is, we can find ๐ ′ . ๐ ′ (๐ฅ) = 6๐ฅ − 2 ๐ ′ (1) = 4 ๐′ (6) = Class Notes 3.4 1 4 Page 9 of 19 √3 Example: If ๐(๐ฅ) = sin ๐ฅ, and f and g are inverse functions, find ๐′ ( 2 ). 1 ๐′ (๐ฅ) = ′ ๐ (๐(๐ฅ)) √3 The expression ๐′ ( 2 ) asks the question what value of x has ๐(๐ฅ) = and trig, we know that sin ๐/3= √3 . 2 From our knowledge of the unit circle √3 . 2 ๐′ ( 1 1 √3 = ′ )= 2 ๐ ′ (๐(√3⁄2)) ๐ (๐/3) 1 Since ๐(๐ฅ) = sin ๐ฅ, ๐ ′ (๐ฅ) = cos ๐ฅ. Therefore, ๐ ′ (๐/3) = cos(๐⁄3) = 2. That means that ๐′ ( 1 1 1 1 √3 = ′ = = =2 )= 2 ๐ ′ (๐(√3⁄2)) ๐ (๐/3) cos(๐⁄3) 1/2 Find ๐′(4) when ๐(๐ฅ) = ๐ฅ 3 − 4 and g is the inverse of f. Find ๐′ (2) when ๐(๐ฅ) = 2๐ฅ 2 − 7๐ฅ + 2 Find ๐′ (1) when ๐(๐ฅ) = √๐ฅ − 3 End of 3.8 Day 1 Class Notes 3.4 Page 10 of 19 3.8 Day 2 Inverse Trig Functions Finding the Derivative of the Inverse Numerically Before we go onto the formulas for the inverse trig functions, let’s revisit yesterday’s work. Using the following table, please find ๐′ (5) when f and g are inverse functions. x ๐(๐ฅ) ๐ ′ (๐ฅ) -3 2 -1 -1 3 1 0 -2 3 1 0 12 2 5 -2 3 8 0 Find ๐′ (8) Find ๐′ (8) Yesterday was about finding the value of the inverse of the derivative of an inverse function. We saw that we could find that value regardless of whether or not we could name the inverse function. Today, we’re going to Now, we’re going to learn the derivatives of the inverse trig functions. You must memorize these.Tomorrow, we’ll derive each one (or you can do the derivation in the book). ๐ ๐ข′ arcsin ๐ข = ๐๐ฅ √1 − ๐ข2 ๐ ๐ข′ arccos ๐ข = − ๐๐ฅ √1 − ๐ข2 ๐ ๐ข′ arctan ๐ข = ๐๐ฅ 1 + ๐ข2 We’ll apply these as we’d do for any other derivative, including the use of our rules (constant multiple, sum and difference, product, quotient, chain, and trig formulae). Differentiate each of the following Ready. Set. Go. arcsin(2๐ฅ) Class Notes 3.4 arctan(3๐ฅ) arcsin √๐ฅ Page 11 of 19 2 arcsin(x − 1) arcsin t 2 arcsin(3๐ฅ) ๐ฅ x 2 arctan(x) Class Notes 3.4 3arccos ๐ฅ 2 arcsin √๐ฅ Page 12 of 19 AP Calculus AB Class Notes 3.9 Derivatives of Exponential and Logarithmic Functions pp 172 - 180 Objectives: Find the derivatives of exponential and logarithmic functions. Use all previous rules of differentiation (constant multiple, power, product, quotient, chain, implicit, trig, inverse) in conjunction with the new rules for logarithms and exponential functions. Outline Differentiate exponential functions. Differentiate logarithmic functions. Vocabulary Omissions Logarithmic differentiation is not part of the AP Calculus AB curriculum. 3.9 Day 1 Exponential Differentiation The derivative of the exponential function is truly the most bizzare in all of calculus. ๐ ๐ฅ ๐ = ๐๐ฅ ๐๐ฅ What does this mean? The slope of the tangent to the function ๐ ๐ฅ at the point x is ๐ ๐ฅ ! Take a look at the table: x 0 1 2 โฎ n ๐(๐ฅ) ๐ 0 or 1 ๐ 0 or ๐ ๐2 โฎ ๐๐ ๐ ′ (๐ฅ) 1 ๐ ๐2 โฎ ๐๐ tangent line ๐ฆ = 1(๐ฅ − 0) + 1 ๐ฆ = ๐(๐ฅ − 1) + ๐ ๐ฆ = ๐ 2 (๐ฅ − 1) + ๐ 2 โฎ ๐ฆ = ๐ ๐ (๐ฅ − ๐) + ๐ ๐ y ๏ฑ๏ฐ ๏ธ ๏ถ ๏ด ๏ฒ x ๏ฑ Class Notes 3.4 ๏ฒ ๏ณ Page 13 of 19 If the argument of ๐ ๐ฅ is a function, for example, ๐ sin ๐ฅ , then the derivative of ๐ ๐ข is given by ๐ ๐ข ๐ = ๐ข′ ๐ ๐ข ๐๐ฅ ๐ Let’s give it a try. Find ๐๐ฅ of the following expressions. ๐ 7๐ฅ ๐ −5๐ฅ ๐ฅ3 ๐๐ฅ ๐ 1/๐ฅ ๐ ๐ฅ − ๐ −๐ฅ ๐ ๐ฅ + ๐ −๐ฅ sin(๐ ๐ฅ ) ๐ ๐ฅ๐ก๐๐ ๐ฅ ๐ (๐ฅ+๐ฅ Class Notes 3.4 2) Page 14 of 19 There is a more obscure exponential function that shows up from time to time. It’s the exponential function in the form of ๐๐ข as in ๐(๐ฅ) = 8๐ฅ . There are two way to look at this. 1) Write the function as a function as an exponential to the base e. Original function. ๐ฆ = 8๐ฅ Take the natural logarithm of the function. Exponentiate to the base e to undo previous step. Differentiate both sides wrt x. Derivative of the exponential function. Bring down the x with the exponent of the argument of a logarithm rule. Bring out the constant with the constant multiple rule. ๐ ๐ฅ=1 ๐๐ฅ Done! Now, let’s create the general form: we do this by using the definition of the logarithm. Original function. ๐ฆ = ๐๐ข Take the natural logarithm of the function. Exponentiate to the base e to undo previous step. Differentiate both sides wrt x. Derivative of the exponential function. Bring down the x with the exponent of the argument of a logarithm rule. Bring out the constant with the constant multiple rule. ๐ ๐ฅ = ๐ข′ ๐๐ฅ Exponent of the logarithm rule. Done! This is shown on page 173 of your book with the general form of the equation: ๐ ๐ข ๐๐ข (๐ ) = ๐๐ข ln ๐ ๐๐ฅ ๐๐ฅ Seen next to it’s cousin, it starts to make sense: Class Notes 3.4 Page 15 of 19 ๐ ๐ข ๐ ๐๐ฅ Try some. 5๐ฅ ๐๐ข ๐ (๐๐ข ) ๐๐ฅ = ๐ ๐ข ๐๐ฅ ๐๐ข = ๐๐ข ln ๐ ๐๐ฅ ๐ Find ๐๐ฅ of the following expressions. 3−๐ฅ ๐ ๐ฅ tan ๐ฅ ๐ sin ๐ฅ End of 3.9 Day 1 3.9 Day 2 Logarithmic Differentiation Since we know that the inverse of the logatrithmic function is the exponential function , and we know how to differentiate the exponential function, we can use our knowledge to find the derivative of ln ๐ข. 1. ๐ฆ = ln ๐ข Original equation. 2. Exponentiate both sides to the base e. 3. Since ๐ ln ๐ข = ๐ข … 4. Reflexive property. 5. Take derivative of both sides wrt x. 6. ๐๐ข Rename ๐๐ฅ . 7. Derivative of exponent rule. 8. From step 4: ๐ ๐ฆ = ๐ข 9. Reflexive property. 10. Solving for ๐ฆ′ we have…. Thus, we have the formula: Let’s try some: ln 2 Class Notes 3.4 ๐ 1 ๐๐ข ln ๐ข = ๐๐ฅ ๐ข ๐๐ฅ ln(๐ฅ 3 ) Page 16 of 19 (ln ๐ฅ)2 ln (sin ๐ฅ) ln(2 + √๐ฅ) ln ( ln √๐ฅ √1 + ln2 ๐ฅ x ) 1 + x2 Logarithims other than base e To find the derivative of log ๐ ๐ข, we’ll use the rule for finding the logarithm with an arbitrary base ln ๐ข log ๐ ๐ข = . ln ๐ Class Notes 3.4 Page 17 of 19 Original problem. ๐ฆ = log ๐ ๐ข Rewrite with the expansion of a logarithm with arbitrary base, b. Take derivative of each side wrt x. Since ln ๐ is a constant, we can take it out with the constant multiple rule. ๐ Recalling from the previous derivation that ๐๐ฅ ln ๐ข = ๐ข′ , ๐ข we get… Our two formulas for the derivative of a logarithm are therefore ๐ 1 ๐๐ข ln ๐ข = ๐ข ๐๐ฅ ๐๐ฅ ๐ Once again, let’s try this on. log 3 ๐ฅ log10 cos ๐ฅ log 5 √๐ฅ 2 − 1 log 2 Class Notes 3.4 1 log ๐ ๐ข = ln ๐ โ ๐๐ฅ and ๐ข′ ๐ข ๐ฅ2 ๐ฅ−1 Page 18 of 19 Assignments simple harmonic motion jerk (p. 143) (p. 144) Chain Rule Power Chain Rule orthogonal curves implicit differentiation normal to the surface Power Rule for Rational Powers of x inverse function–inverse cofunction identities Power Rule for Arbitrary Real Powers logarithmic differentiation orthogonal families derivative of f at a differentiable function (p. 149) (p. 151) (p. 154) (p. 157) (p. 159) (p. 161) (p. 168) (p. 176) (p. 177) (p. 180) (p. 99) (p. 99) Class Notes 3.4 Page 19 of 19