The Explicit Form of a Function
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1 Section 3 5 Implicit Differentiation The Eplicit Form of a Function Function Notation requires that we state a function with f () on one sie of an equation an an epression in terms of on the other sie of the equation. 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 () = f () = sin The Implicit Form of a Function 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 of a Relation. Eample 1 Eample 2 y = 6 y +1 2 = 6 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 () = Many equations state in implicit form are functions but it may be ifficult to solve for y. In this case you may not know if the relation is a function unless you can analyze the graph of the relation. For the purposes of this unit we o not nee to etermine if the relation is a function. We will keep each equation in Implicit Form an fin ways to ifferentiate the epression irectly form itʼs implicit form. 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 factor that remains The erivative of with respect to is 1 = ( ) =1 The erivative of y with respect to is ( y) = y = The Chain Rule for the erivative of u n with respect to u u un = n u n 1 u The Chain Rlue for the eriavtive of 3 with respect to 3 = 3 2 = =1 3 = 3 The Chain Rlue for the eriavtive of y 3 with respect to y3 = 3 y 3 y = y3 = 3 y 2 Math Page 2 of Eitel
3 Many equations are not functions an cannot be states in eplicit form as y = f() Some functions are too ifficult to solve for y. In either case these relations are state in implicit form. To ifferentiate these functions we use Implicit ifferentiation. Implicit ifferentiation. Step 1: Differentiate both sie of the equation with respect to. Step 2: Move all terms with a factor tn the left sie of the equation an move the remining terms to the left sie. Step 3: Factor out the common factor from each term on the left sie. Step 4: Solve for by iviing by the epression in the left that is multiplie by Eample 1 Fin y given y 4 3y = Differerentiate both sies of the equation with respect to 4u u factor out the = y 3 3 sovle for y 3 3 Fin the slope of the tangent line at (0,0) at (0,0) is 2(0) (0) 3 3 = 3 5 Math Page 3 of Eitel
4 Eample 2 Fin y given y 2y 3 = 2 3 Differerentiate both sies of the equation with respect to the 67 first 8 er 67 of 8 sec 6 the 7 sec 8 + y er 67 of 8 first 1 2y y 2y get all the terms on one sie 2y y factor out a = 6 2 y 2y 2 sovle for 62 y 2y 2 Math Page 4 of Eitel
5 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 1 y value. For this reason both the an y values must be given when the slope is require. Plug the values of an y into the epression for 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 terms alone on one sie 8y 2 sovle for 2 8y = 4y is The slope of the tangent line at the point 1,2 4y = 1 8 Math Page 5 of Eitel
6 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 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 terms alone on one sie 4y 2 sovle for 2 4y = 2y is The slope of the tangent line at the point 3, 1 4y = ( 3) 4( 1) = 3 4 The equation of the tangent line at the point ( 3, 1 ) is y +1 = ) Math Page 6 of Eitel
7 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 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 = simplify (a the fractions) y = 6 4 ( 4) multiply both sies by y 6 4 y ( 4) substitute y = ( 4) 5 in for y ( 4) ( 4) 4 ( 6 4) Math Page 7 of Eitel
8 Fin 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 = simplify (a the fractions) y = 3 ( 3) multiply both sies by y 3 y ( 3) substitute y = ( 3) in for y ( 3) 2 Math Page 8 of Eitel
9 Fin 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 = simplify (a the fractions) y = + 4 ( + 2) multiply both sies by y + 4 y ( + 2) substitute y = 2 in for y ( + 2) Math Page 9 of Eitel
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