The Explicit Form of a Function

Transcription

1 Section 3 5 Implicit Differentiation The Eplicit Form of a Function The normal way we see function notation has f () on one sie of an equation an an epression in terms of on the other sie. We know the equation represents a function because the f () notation eplicitly states that the epression in terms of represents a function. This form is calle the eplicit form. Eample 1 Eample 2 f () = The Implicit Form of a Function f () = sin Some functions are only implie by an equation. In this case, there is not an f () epression on one sie of the equation. The an y variables can appear on either sie of the equation an can also be in the same term. This form is calle the Implicit Form. You CANNOT conclue the relation is a function. One way to etermine if the relation is a function is to try an solve for y to get the epression in Eplicit Form. Eample 1 Eample 2 y =1 solve for y y = 1 This is the graph of an inverse relation so the relation is a function f () = 1 y +1 2 = 6 solve for y y = This is the graph of a parabola so the relation is a function f () = Solving for y to get the eplicit form is not always possible. Many equations state in implicit form are functions but it may be ifficult to solve for y. 2 2y 3 + y = 9 Many equations are epresse in implicit form an are relations an not a function. Circles, hyperbolas an ellipses are eamples of this type of relation. The stanar form for many relations is an implicit form. Circle 2 + y 2 = 9 Ellipse 2 + 4y 2 = 9 Hyperbola 2 4y 2 = 9 The slope of a function at any given value of can be foun by fining the value of the erivative for the given value of. The question of how to fin the erivative of a function or a relation when we cannot (or chose not to) solve for y is the topic of this section. Math Page 1 of Eitel

2 To unerstan how to fin y Implicit Differentiation implicitly we must review what taking the erivative with respect to means. When you ifferentiate with respect to you o not nee to use the chain rule. This is because the factor in the chain rule is equal to 1. When you ifferentiate y with respect to the chain rule has a y factor that remains y The erivative of with respect to is 1 = ( ) =1 The erivative of y with respect to is ( y) = y = y The Chain Rule for the erivative of (u) n states that If y = (u) n then n u n 1 u The Chain Rule for the erivative of 3 with respect to 3 = 3 2 = =1 3 = = 3 2 The Chain Rule for the erivative of y 3 with respect to y3 = 3 y 2 y y = y y3 = 3 y 2 y Math Page 2 of Eitel

3 Implicit ifferentiation with the Chain Rule The Chain Rule for the erivative of (u) n states that (u)n = n u n 1 u Eample 1 The Chain Rule for the erivative of (y) 3 with respect to ( n} 6 u7 n 1 8 y } (y) )3 = 3 (y) 3 1 y ( y)3 = 3 (y) 2 y y3 = 3y 2 y Eample 2 The Chain Rule for the erivative of (y) 1/2 with respect to n} ( y)1/2 = 1 2 u n (y) 1/2 1 } (y) y n} ( y)1/2 = 1 2 u n (y) 1/2 } (y) y ( y)1/2 = y 1/2 y y 1/2 = y 2 y Math Page 3 of Eitel

4 sin(u) [ ] = cos(u) u Eample 3 u } an (u)n = n u n 1 u The Chain Rule for the erivative of sin ( y 2 ) with respect to sin y2 6 cos(u) 78 = cos( y 2 ) 67 u 8 y2 an y2 = 2 y y so sin y2 = 2 y cos( y 2 ) y Eample 4 (u)n = n u n 1 u an [ sin(u) ] = sin(u) [ u] The Chain Rule for the erivative of [ sin(y) ] 2 with respect to [ sin(y) ]2 = 2 an sin(y) so n} u n 1 sin(y) [ ] = cos(y) y sin(y) [ sin(y) ]2 = 2sin(y) cos(y) y The ouble angle rule forsin 2sin(y) cos(y) = sin(2y) so [ sin(y) ]2 = sin(2y) y [ ] Math Page 4 of Eitel

5 Eample 5 ln u [ ] = u The Chain Rule for the erivative of ln( 3y) with respect to u ln 3y ln 3y u u = = 3y 3y ln 3y = 3 y 3y ln 3y = y y Implicit ifferentiation with the prouct rule The Prouct Rule for the erivative of u v states that the = } first u u v er 64 of 7 secon 48 v Eample 1 the 6 secon 78 + v er 6 of 78 first u The Prouct Rule for the eriavtive of y with respect to the = } first y er 64 of 7 secon 48 y the 6 secon 78 + y er 6 of 78 first 1 ( y) = y + y Math Page 5 of Eitel

6 Eample 2 The Prouct Rule for the eriavtive of y 2 with respect to the } first = y2 er 6 of 7 secon 8 y2 the } secon + y 2 er 6 of 78 first the } first = y2 6 er 47 of 4 y y 1 y the } secon + y 2 er of } first 1 ( ) y2 = 2 y y + y 2 Eample 2 The Prouct Rule for the eriavtive of y 2 with respect to sin y = cos y y the = } first y er of } secon y the secon } + y er 6 of 78 first the = } first y er of } y 2 y the secon } + y er of } first 1 ( ) y2 = y + y sin( y) = cos( y) ( y + y) Math Page 6 of Eitel

7 Implicit ifferentiation. Step 1: Differentiate both sies of the equation with respect to. The erivative of will be 1 an the erivative of y will be y. Step 2: Move all terms with a y factor to one sie of the equation an move the remining terms to the other sie. Step 3: Factor out the common y factor from each term on the sie where all the terms have a y. Step 4: Solve for y by iviing both sie of the equation by the epression that is multiplie by y Eample 1 Fin y given y 4 3y = Differerentiate both sies of the equation with respect to 64 4u7 3 4 u 8 4 y 3 y the y terms are alreay alone on the same sie of the equation 4y 3 y factor out the y = y 4y 3 3 sovle for y y 3 3 Math Page 7 of Eitel

8 Eample 2 Fin y given y y 2 = 3 2 Differerentiate both sies of the equation with respect to the 67 first 8 er 67 of 8 sec y 6 the 7 sec 8 + y er 67 of 8 first 1 62u u 2 y y = 6 1 the y terms are alreay alone on the same sie of the equation y + y 2y 6 1 factor out the y = 6 1 y 2y sovle for y 6 1 2y Math Page 8 of Eitel

9 Eample 3 Fin y given y = y2 + Differerentiate both sies of the equation with respect to the 67 first 8 er 67 of 8 sec y 6 the 7 sec 8 + y er 67 of 8 first 1 62u u = 2 y y er 67 of y + y = 2 y y +1 get all the y terms on one sie y 2y 1 y factor out a y =1 y y 2y sovle for y 1 y 2y Math Page 9 of Eitel

10 Eample 4 Fin y given y 2 = y Differerentiate both sies of the equation with respect to the 67 first 8 6 er 47 of 4 sec 8 y 2 the 67 sec 8 + y 2 y y + y = y er 47 of first 48 = er of y y + 6 er 47 of get all the y terms on one sie 2 y y 6 2 y factor out a y y ( 2y 1 ) = 6 2 y sovle for y 62 y 2y 1 Math Page 10 of Eitel

11 Fining the slope of the line tangent to the graph at a point (, y ) The implicit equation may not be a function. You cannot assume that every value has eactly one y value. If there are more than one y value for a given then both the an y values must be given when the slope is require. Plug the values of an y into the epression for y to fin the slope of the line tangent to the graph of the relation at a given point (,y) Given the ellipse ( 1, 2 ) 2 + 4y 2 =17 when = 1 the value(s) for y are y = ±2 ( 1, 2 ) both (1,2) an (1, 2) are points on the graph Eample 3 Fin the slope of the tangent line fo 2 + 4y 2 =17 at the point ( 1,2 ) Differerentiate both sies of the equation with respect to 2 + 8y 0 get all the y terms alone on one sie 8y 2 sovle for y 2 8y = 4y is The slope of the tangent line at the point 1,2 4y = 1 8 Math Page 11 of Eitel

12 Fining the equation of the line tangent to the graph at a point (, y ) The equation of the line tangent to the graph at the point 1, y 1 is given by the equation y y 1 = m( ( 1 ) where m is the slope of the tangent line an m is the value of y at 1, y 1 Given the ellipse 2 + 2y 2 =11 the point ( 3, 1) is a point on the graph ( 3, 1 ) Eample 4 Fin the slope of the tangent line fo 2 + 2y 2 =11 at the point ( 3, 1 ) Differerentiate both sies of the equation with respect to 2 + 4y 0 get all the y terms alone on one sie 4y 2 sovle for y 2 4y = 2y is The slope of the tangent line at the point 3, 1 2y = ( 3) 2( 1) = 3 2 = 3 2 The equation of the tangent line at the point ( 3, 1 ) is y +1 = Math Page 12 of Eitel

13 Logarithmic Differentiation If y = f () where f () > 0 then you can take the natural logarithm of both sies of the equation ln(y) = ln f () This allows us to use the properties of logs to simplify the epression for f (). The properties of logs change proucts into sums, fractions into ifferences an powers into proucts. After we simplify the logarithmic epression we can then ifferentiate implicitly. Eample 1 Fin y if y = ( 4) 5 take the natural log of both sies of the equation [ ] ln (y) = ln ( 4) 5 ln (y) = ln() + 5ln ( 4 ) take the erivative of both sies of the equation y y = simplify (a the fractions) y y = 6 4 ( 4) = ( 4) multiply both sies by y y ( 4) substitute y = ( 4) 5 in for y 5 2( 3 2) 4 ( 4) 2( 4) 4 ( 3 2) Math Page 13 of Eitel

14 Fin y if y = Eample 2 3 take the natural log of both sies of the equation ln (y) = ln 3 ln (y) = ln() ln ( 3) take the erivative of both sies of the equation y y = simplify (a the fractions) y y = 3 ( 3) multiply both sies by y 3 y ( 3) substitute y = ( 3) in for y ( 3) 2 Math Page 14 of Eitel

15 Fin y if y = Eample 3 take the natural log of both sies of the equation ln (y) = ln ln (y) = 2ln() ln ( + 2) take the erivative of both sies of the equation y y = simplify (a the fractions) y y = + 4 ( + 2) multiply both sies by y + 4 y ( + 2) substitute y = 2 in for y ( + 2) Math Page 15 of Eitel