Implicit Differentiation

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Implicit Differentiation Implicit Differentiation Using the Chain Rule In the previous section we focuse on the erivatives of composites an saw that THEOREM 20 (Chain Rule) Suppose that u = g(x) is ifferentiable at x an y = f (x) is ifferentiable at u = g(x) Then the composite f (g(x)) is ifferentiable at x an This can be written as y x = y u u x x [ f (g(x))] = f (g(x))g (x) Notice that we can apply the chain rule even if we o not know the precise formula for the inner function EXAMPLE 202 Write out the chain rule erivative for each these, assuming u an v are functions of x u (a) (6 sin(u)) = 6 cos u Note the since u is some unknown function of x, we must x x inclue the chain rule erivative ( insie erivative ), u/x (b) x [eu ] = e u u x u (e) [sec u] = sec u tan u x x ( ) (g) u 3 cos(v) x (c) (f ) u [sin u] = cos u x x x [un n u ] = nu x () x [tan u] = sec2 u u x = 3u 2 u x cos(v) + u3 ( sin(v)) v x = 3u2 cos(v) u x u3 sin(v) v x Notice that the symbol (letter) we use for the inner function oes not matter For example, if we use y for the insie function instea of u, we have y = g(x), then Similarly y 3 {}}{ x [ (g(x)) 3 ] = 3y 2 y x = 3(g(x))2 g (x) or x (y3 ) = 3y 2 y x More generally, you text states y 5 {}}{ x [ (g(x)) 5 ] = 5u 4 y x = 5(g(x))4 g (x) or x (y5 ) = 5y 4 y x Now what we nee to pay attention to is the variables use in the function an the variable that the erivative is taken with respect to Notice that THEOREM 203 (General Power Rule) If f (x) is a ifferentiable function, then x [( f (x))n ] = n( f (x)) n f (x) x (u9 ) = 9u 8 u x

2 Similarly x (y9 ) = 9y 8 y x an t (x9 ) = 9x 8 x t However x (x9 ) = 9x 8, because the variables are the same an there is no chain rule insie erivative EXAMPLE 204 Here are several other examples: () (3) x (tan u) = sec2 u u (2) x x (tan y) = sec2 y y x x (tan x) = sec2 x (4) t (a2 ) = 2a a t 5 If c 2 = a 2 + b 2, then t (c2 ) = t (a2 + b 2 ) so 2c c a b = 2a + 2b t t t 6 If that A = πr 2, then t (A) = A t (πr2 ) so = 2πr r Here we are assuming that t t the raius of the circle is changing with respect to time In other wors, we assume that r is some (unknown) function of t EXAMPLE 205 Fancier examples: This time assume u, v, an y are functions of x Using the prouct rule, 2 An x (x2 y 3 ) = 2xy 3 + x 2 3y 2 y x Implicit an Explicit Functions x (u2 v 3 ) = 2u u x v3 + u 2 3v 2 v x Sometimes we have an explicit formula for a function For example, y = f (x) = e x or y = f (x) = x 2 + 3x + efines y explicitly as a function of x For each value of x that we put into either function, we obtain a single output value Compare this to the equation xy 2 = Here the relation between x an y is efine implicitly Rather than a function we have an equation involving both x an y In this case we coul solve for y an obtain y 2 = x which implies y = x or y = x The result is not a function We will take as our efinition that an implicit function is an equation involving two or more variables EXAMPLE 206 Consier the implicit function given by x 2 + y 2 = We know that its graph is a circle an is not an orinary function since its graph oes not pass the vertical line test In this case, we coul create two orinary (explicit) functions from this single implicit function Solving for y we fin that y 2 = x 2 so that y = x 2 or y = x 2 EXAMPLE 207 Here s another implicitly efine function: x 3 + yx 2 + y 3 = 7 This time there is no simple way to make one of more explicit functions from this relation Figure 20: The graph of the relation xy 2 = is not the graph of a function It oes not pass the vertical line test Figure 202: The graph of the relation x 2 + y 2 = is not the graph of a function It oes not pass the vertical line test The Goal Given an implicit function in variables x an y, fin the erivative y without ever x solving for y explicitly This is calle implicit ifferentiation The next example illustrates how to o this

3 EXAMPLE 208 Determine y x if x2 + y 2 = Then calculate the slope of the curve at the points ( 2, 3 2 ), (0, ), an (, 0) SOLUTION Note first that since we want to etermine y x, we are treating x as the inepenent variable So we are taking the erivative with respect to x Thus, x (x2 + y 2 ) = x () so x (x2 ) + x (y2 ) = 0 Using the chain rule 2x + 2y y x = 0 Now solve for y x so 2y y x = 2x y x = x, (y = 0) y Now we can etermine slopes at specific points on the circle The slope at ( 2, 3 2 ) check that this point is on the circle is given by substituting in the particular x an y values of the point into the expression for y x Notice the notation use to inicate the point at which the erivative is calculate y x = x ( 2, 3 2 ) y = ( 2, 3 2 ) At (0, ) 2 = = 3 3 2 y x = x (0,) y = 0 (0,) = 0 3 3 Looking at the graph of the circle, these two slopes look to be correct Finally, the slope at (, 0) is not efine since y x = x, (y = 0) The tangent line is vertical y EXAMPLE 209 (Like an exam question) Let x 2 + 2xy = y 3 + 3 Fin y x 2 Verify that (3, 2) is on the curve 3 Determine the equation of the tangent line at (3, 2) SOLUTION () Take the erivative (with respect to x) of both sies of the equation x (x2 + 2xy) = x (y3 + 3) Using the prouct an chain rules 2x + 2y + 2x y y = 0 + 3y2 x x Gather all the y x terms on the left sie Solve for y x 2x y y 3y2 = 2x 2y x x (2x 3y 2 ) y = 2x 2y x y x = 2x 2y 2x 3y 2 Figure 203: The tangents at ( 2, 3 2 ) an (0, ) The graph of x 2 + 2xy = y 3 + 3

4 2 Check that (3, 2) is on the curve: x 2 + 2xy = 3 2 + 2(3)(2) = 2 an 3 + y 3 = 3 + 2 3 = 2 Thus (3, 2) is on the curve 3 Determine the tangent at (3, 2) The slope is m = y x = (3,2) 2x 2y 2x 3y 2 = (3,2) 2(3) 2(2) 2(3) 3(2 2 ) = 0 6 = 5 3 y 2 = 5 3 (x 3) or y = 5 3 x 5 + 2 or y = 5 3 x 3 EXAMPLE 200 (Like an exam question) Let x 2 + x 2 y 2 = y 3 3 Verify that (, 2) is on the curve 2 Determine the equation of the tangent line at (, 2) SOLUTION Check that (, 2) is on the curve At (, 2) x 2 + x 2 y 2 = 2 + 2 2 2 = 5 an y 3 3 = 2 3 3 = 5 Thus (, 2) is on the curve 2 To fin the tangent at (, 2) we nee the slope; we alreay have the point We nee to fin y x Take the erivative (with respect to x) of both sies of the equation x (x2 + x 2 y 2 ) = x (y3 + 3) 2x + 2xy 2 + 2x 2 y y y = 3y2 x x Gather all the y x terms on the left sie (2x 2 y 3y 2 ) y = 2x 2xy2 x Solve for y x y 2x 2xy2 = x 2x 2 y 3y 2 Evaluate y x at (, 2) to get the slope m = y 2x 2xy2 x = (,2) 2x 2 y 3y 2 = 2() 2()(22 ) (,2) 2( 2 )(2) 3(2 2 ) = 0 8 = 5 4 y 2 = 5 4 (x ) or y = 5 4 x 5 4 + 2 or y = 5 4 x + 3 4 EXAMPLE 20 (Easier) Let x + y 2 4y = 2 Determine the equations of all the tangents when x = 2 2 At what points are the tangents vertical SOLUTION First we must etermine the y-coorinates of the points on the curve that have x = 2 as the first coorinate Since x + y 2 4y = 2, when x = 2 2 + y 2 4y = 2 so y 2 4y = y(y 4) = 0 or y = 0, 4 So the points are (2, 0) an (2, 4)

5 2 To fin the tangents we nee the slope so we nee to fin y x Take the erivative (with respect to x) of both sies of the equation Solve for y x x (x + y2 4y) = x (2) + 2y y x 4 y x = 0 (2y 4) y x = y x = 2y 4 (y = 2) At (2, 0): Evaluate y at (2, 0) to get the slope at the first point x m = y x = (2,0) 2y 4 = (2,0) 2(0) 4 = 4 y 0 = 4 (x 2) or y = 4 x 2 See Figure 204 At (2, 4): m = y x = (2,4) 2y 4 = (2,4) 2(4) 4 = 4 See Figure 204 y 4 = 4 (x 2) or y = 4 x 9 2 3 The graph will have a vertical tangent where the slope y x = 2y 4 becomes infinite This is where the enominator goes to 0 This is at y = 2 When y = 2, x + y 2 4y = 2 so x + 2 2 4(2) = 2 so x = 2 4 + 8 = 6 Thus, there is a vertical tangent at (6, 2) See Figure 204 EXAMPLE 202 Fin y x if cos y + sin x = x2 SOLUTION Take the erivative with respect to x an solve for y x EXAMPLE 203 Fin y x if exy = x (cos y + sin x) = x x (x2 ) sin y y + cos x = 2x x sin y y = 2x cos x x y 2x cos x = = x sin y cos x 2x sin y SOLUTION Take the erivative with respect to x of both sies of the equation an solve for y x x (exy ) = x (x) ( e xy y + x y ) = x e xy y + e xy x y x = e xy x y x = exy y y x = exy y e xy x (2, 4) (2, 0) Figure 204: The graph of the relationx + y 2 4y = 2 (6, 2)

6 EXAMPLE 204 Fin y x if tan(xy) = x2 + y SOLUTION Take the erivative with respect to x of both sies of the equation an solve for y x x (tan(x3 y 3 )) = x (x2 + y) ( sec 2 (xy) 3x 2 y 3 + 3x 3 y 2 y ) = 2x + y x x 3x 2 y 3 sec 2 (xy) + 3x 3 y 2 sec 2 (xy) y y = 2x + x x 3x 3 y 2 sec 2 (xy) y x y x = 2x 3x2 y 3 sec 2 (xy) (3x 3 y 2 sec 2 (xy) ) y x = 2x 3x2 y 3 sec 2 (xy) YOU TRY IT 20 Determine y x given ex+2y = x y x = 2x 3x2 y 3 sec 2 (xy) 3x 3 y 2 sec 2 (xy) YOU TRY IT 202 Determine y x given ey + 9y = 4x 5 + y x = 2e x+2y 2 answer to you try it 20 y x = 20x4 e y + 9 answer to you try it 202 YOU TRY IT 203 Determine the equation of the tangent line to y x given ey + 9y = 4x 5 + at (, 2) answer to you try it?? y x = y3 2xy 3xy 2 + x 2 Tangent: y 2 = 3 5 (x )

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