Preface. Implicit Differentiation. y =

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1 Preface Here are the solutions to the practice problems for m Calculus I notes. Some solutions will have more or less detail than other solutions. The level of detail in each solution will depend up on several issues. If the section is a review section, this mostl applies to problems in the first chapter, there will probabl not be as much detail to the solutions given that the problems reall should be review. As the difficult level of the problems increases less detail will go into the basics of the solution under the assumption that if ou ve reached the level of working the harder problems then ou will probabl alread understand the basics fairl well and won t need all the eplanation. This document was written with presentation on the web in mind. On the web most solutions are broken down into steps and man of the steps have hints. Each hint on the web is given as a popup however in this document the are listed prior to each step. Also, on the web each step can be viewed individuall b clicking on links while in this document the are all showing. Also, there are liable to be some formatting parts in this document intended for help in generating the web pages that haven t been removed here. These issues ma make the solutions a little difficult to follow at times, but the should still be readable. Implicit Differentiation 1. For = 1 do each of the following. (a) Find b solving the equation for and differentiating directl. (b) Find b implicit differentiation. (c) Check that the derivatives in (a) and (b) are the same. (a) Find b solving the equation for and differentiating directl. First we just need to solve the equation for. Step Now differentiate with respect to. 1 = = = 1 (b) Find b implicit differentiation. 007 Paul Dawkins 1

2 Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! Also, don t forget that because is reall ( ) we ma well have a Product and/or a Quotient Rule buried in the problem. First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. Also, prior to taking the derivative a little rewrite might make this a little easier. = 1 Now take the derivative and don t forget that we actuall have a product of functions of here and so we ll need to use the Product Rule when differentiating the left side. 4 = 0 Step Finall all we need to do is solve this for. = = 4 (c) Check that the derivatives in (a) and (b) are the same. Hint : To show the are the same all we need is to plug the formula for (which we alread have.) into the derivative we found in (b) and, potentiall with a little work, show that we get the same derivative as we got in (a). From (a) we have a formula for written eplicitl as a function of so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a). So, we got the same derivative as we should. 1 = = = 1. For + = 4 do each of the following. (d) Find b solving the equation for and differentiating directl. (e) Find b implicit differentiation. (f) Check that the derivatives in (a) and (b) are the same. 007 Paul Dawkins

3 (a) Find b solving the equation for and differentiating directl. First we just need to solve the equation for. Step Now differentiate with respect to. ( ) 1 = 4 = 4 ( ) = 4 (b) Find b implicit differentiation. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. Taking the derivative gives, Step Finall all we need to do is solve this for. + = 0 = (c) Check that the derivatives in (a) and (b) are the same. Hint : To show the are the same all we need is to plug the formula for (which we alread have.) into the derivative we found in (b) and, potentiall with a little work, show that we get the same derivative as we got in (a). From (a) we have a formula for written eplicitl as a function of so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a). So, we got the same derivative as we should. = = = 4 ( ) ( ) Paul Dawkins

4 . For + = do each of the following. (g) Find b solving the equation for and differentiating directl. (h) Find b implicit differentiation. (i) Check that the derivatives in (a) and (b) are the same. (a) Find b solving the equation for and differentiating directl. First we just need to solve the equation for. ( ) 1 = =± Note that because we have no restriction on (i.e. we don t know if is positive or negative) we reall do need to have the ± there and that does lead to issues when taking the derivative. Hint : Two formulas for and so two derivatives. Step Now, because there are two formulas for we will also have two formulas for the derivative, one for each formula for. The derivatives are then, 1 1 ( ) ( ) ( 0) = = > 1 1 ( ) ( ) ( 0) = = < As noted above the first derivative will hold for > 0 while the second will hold for < 0 and we can use either for = 0 as the plus/minus won t affect that case. (b) Find b implicit differentiation. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. Taking the derivative gives, + = 0 Step Finall all we need to do is solve this for. 007 Paul Dawkins 4

5 = (c) Check that the derivatives in (a) and (b) are the same. Hint : To show the are the same all we need is to plug the formula for (which we alread have.) into the derivative we found in (b) and, potentiall with a little work, show that we get the same derivative as we got in (a). Again, two formulas for so two derivatives From (a) we have a formula for written eplicitl as a function of so plug that into the derivative we found in (b) and, with a little simplification/work, show that we get the same derivative as we got in (a). Also, because we have two formulas for we will have two formulas for the derivative. First, if > 0 we will have, = = = = ( ) Net, if < 0 we will have, ( ) ( ) =( ) = = = ( ) So, in both cases, we got the same derivative as we should. ( ) 4. Find b implicit differentiation for =. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. Differentiating with respect to gives, = 6 Step Finall all we need to do is solve this for Paul Dawkins 5

6 ( ) = = sin = Find b implicit differentiation for ( ) Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. Differentiating with respect to gives, 14 + cos( ) = 4 Step Finall all we need to do is solve this for. ( 14 4 ) cos( ) ( ) cos + = = Find b implicit differentiation for sin ( ) e =. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. Differentiating with respect to gives, e ( ) cos = 1 Don t forget the on the cosine after differentiating. Again, is reall ( ) and so when differentiating sin ( ) we reall differentiating sin ( ) the Chain Rule! and so we are differentiating using 007 Paul Dawkins 6

7 Step Finall all we need to do is solve this for. 1 e = = cos ( ) ( e 1) sec( ) 7. Find b implicit differentiation for = + 4. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! Also, don t forget that because is reall ( ) we ma well have a Product and/or a Quotient Rule buried in the problem. First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. This also means that the first term on the left side is reall a product of functions of and hence we will need to use the Product Rule when differentiating that term. Differentiating with respect to gives, = Step Finall all we need to do is solve this for. ( ) = 1 8 = Find b implicit differentiation for ( ) cos + + e = 1. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! Also, don t forget that because is reall ( ) we ma well have a Product and/or a Quotient Rule buried in the problem. First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. This also means that the second term on the left side is reall a product of functions of and hence we will need to use the Product Rule when differentiating that term. Differentiating with respect to gives, 007 Paul Dawkins 7

8 ( ) ( ) + sin + + e + e = 0 Step Finall all we need to do is solve this for (with some potentiall mess algebra). ( ) ( ) e e ( e sin ( ) ) 0 sin ( ) sin + sin = 0 + = + + e = ( + ) e ( + ) sin e sin 9. Find b implicit differentiation for ( ) tan 4 = +. Hint : Don t forget that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving! Also, don t forget that because is reall ( ) we ma well have a Product and/or a Quotient Rule buried in the problem. First, we just need to take the derivative of everthing with respect to and we ll need to recall that is reall ( ) and so we ll need to use the Chain Rule when taking the derivative of terms involving. This also means that the when doing Chain Rule on the first tangent on the left side we will need to do Product Rule when differentiating the inside term. Differentiating with respect to gives, 4 4 ( + 4 ) sec ( ) = + Step Finall all we need to do is solve this for (with some potentiall mess algebra). ( ) ( ) ( ) + = sec 4 sec ( 4sec ) = sec ( ) = sec ( ) ( ) sec 10. Find the equation of then tangent line to + = at ( 1, ). 4 Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn t be too hard at this point. 007 Paul Dawkins 8

9 The first thing to do is use implicit differentiation to find for this function. 4 + = 0 = Step Evaluating the derivative at the point in question to get the slope of the tangent line gives, m = = = = 1, = Step Now, we just need to write down the equation of the tangent line. ( ) ( 1) ( 1) ( ) = = = 11. Find the equation of then tangent line to = + e at ( ) 0,. Hint : We know how to compute the slope of tangent lines and with implicit differentiation that shouldn t be too hard at this point. The first thing to do is use implicit differentiation to find for this function. e e e + = + = e Step Evaluating the derivative at the point in question to get the slope of the tangent line gives, m 18 = = = 6 = 0, = Step Now, we just need to write down the equation of the tangent line. =6 0 = 6+ ( ) 1. Assume that = ( t), = ( t) and z z( t) respect to t. = and differentiate 4 + z = 1 with 007 Paul Dawkins 9

10 Hint : This is just implicit differentiation like we ve been doing to this point. The onl difference is that now all the functions are functions of some fourth variable, t. Outside of that there is nothing different between this and the previous problems. Solution Differentiating with respect to t gives, + 4z z = 0 Note that because we were not asked to give the formula for a specific derivative we don t need to go an farther. We could however, if asked, solved this for an of the three derivatives that are present. 1. Assume that = ( t), = ( t) and z z( t) with respect to t. = and differentiate cos( ) = sin ( + 4z) Hint : This is just implicit differentiation like we ve been doing to this point. The onl difference is that now all the functions are functions of some fourth variable, t. Outside of that there is nothing different between this and the previous problems. Solution Differentiating with respect to t gives, ( ) ( ) = ( + ) ( + ) cos sin 4z cos 4z Note that because we were not asked to give the formula for a specific derivative we don t need to go an farther. We could however, if asked, solved this for an of the three derivatives that are present. 007 Paul Dawkins 10