3.6. Implicit Differentiation. Implicitly Defined Functions

Transcription

1 3.6 Implicit Differentiation Implicit Differentiation (3, ) Slope 3 FIGURE 3.36 The circle combines the graphs of two functions. The graph of 2 is the lower semicircle an passes through s3, -. Most of the functions we have ealt with so far have been escribe b an equation of the form ƒs that epresses eplicitl in terms of the variable. We have learne rules for ifferentiating functions efine in this wa. In Section 3.5 we also learne how to fin the erivative > when a curve is efine parametricall b equations st an st. A thir situation occurs when we encounter equations like , 2-0, or (See Figures 3.36, 3.37, an 3.38.) These equations efine an implicit relation between the variables an. In some cases we ma be able to solve such an equation for as an eplicit function (or even several functions) of. When we cannot put an equation Fs, 0 in the form ƒs to ifferentiate it in the usual wa, we ma still be able to fin > b implicit ifferentiation. This consists of ifferentiating both sies of the equation with respect to an then solving the resulting equation for. This section escribes the technique an uses it to eten the Power Rule for ifferentiation to inclue rational eponents. In the eamples an eercises of this section it is alwas assume that the given equation etermines implicitl as a ifferentiable function of. Implicitl Define Functions We begin with an eample.

2 206 Chapter 3: Differentiation 2 Slope 2 0 Slope 2 P(, ) Q(, ) FIGURE 3.37 The equation 2-0, or 2 as it is usuall written, efines two ifferentiable functions of on the interval Ú 0. Eample shows how to fin the erivatives of these functions without solving the equation 2 for ( 0, ) ( 0, 2 ) ( 0, 3 ) f () A f 2 () f 3 () FIGURE 3.38 The curve is not the graph of an one function of. The curve can, however, be ivie into separate arcs that are the graphs of functions of. This particular curve, calle a folium, ates to Descartes in 638. EXAMPLE Fin > if 2. Differentiating Implicitl Solution The equation 2 efines two ifferentiable functions of that we can actuall fin, namel an 2 - (Figure 3.37). We know how to calculate the erivative of each of these for 7 0: But suppose that we knew onl that the equation 2 efine as one or more ifferentiable functions of for 7 0 without knowing eactl what these functions were. Coul we still fin >? The answer is es. To fin >, we simpl ifferentiate both sies of the equation 2 with respect to, treating ƒs as a ifferentiable function of : This one formula gives the erivatives we calculate for both eplicit solutions an 2 -: EXAMPLE 2 2 Slope of a Circle at a Point Fin the slope of circle at the point s3, -. Solution The circle is not the graph of a single function of. Rather it is the combine graphs of two ifferentiable functions, an (Figure 3.36). The point s3, - lies on the graph of 2, so we can fin the slope b calculating eplicitl: 2 ` 3 2 an an ` 3 But we can also solve the problem more easil b ifferentiating the given equation of the circle implicitl with respect to : - A2 B + A2 B A25B The Chain Rule gives A2 B CƒABD2 2ƒsƒ s 2. 2A - B The slope at s3, - is - ` - 3 s3,

3 3.6 Implicit Differentiation sin Notice that unlike the slope formula for 2 >, which applies onl to points below the -ais, the formula > -> applies everwhere the circle has a slope. Notice also that the erivative involves both variables an, not just the inepenent variable. To calculate the erivatives of other implicitl efine functions, we procee as in Eamples an 2: We treat as a ifferentiable implicit function of an appl the usual rules to ifferentiate both sies of the efining equation. EXAMPLE 3 Differentiating Implicitl Fin > if sin (Figure 3.39) FIGURE 3.39 The graph of sin in Eample 3. The eample shows how to fin slopes on this implicitl efine curve. Solution scos a b 2 + scos s2 - cos sin A2 B A2 B + Asin B scos AB 2 + scos a + b 2 + cos 2 + cos 2 - cos Differentiate both sies with respect to Á Á treating as a function of an using the Chain Rule. Treat as a prouct. Collect terms with > Á Á an factor out >. Solve for > b iviing. Notice that the formula for > applies everwhere that the implicitl efine curve has a slope. Notice again that the erivative involves both variables an, not just the inepenent variable. Normal line Light ra A P Tangent Point of entr B Curve of lens surface FIGURE 3.0 The profile of a lens, showing the bening (refraction) of a ra of light as it passes through the lens surface. Implicit Differentiation. Differentiate both sies of the equation with respect to, treating as a ifferentiable function of. 2. Collect the terms with > on one sie of the equation. 3. Solve for >. Lenses, Tangents, an Normal Lines In the law that escribes how light changes irection as it enters a lens, the important angles are the angles the light makes with the line perpenicular to the surface of the lens at the point of entr (angles A an B in Figure 3.0). This line is calle the normal to the surface at the point of entr. In a profile view of a lens like the one in Figure 3.0, the normal is the line perpenicular to the tangent to the profile curve at the point of entr.

4 208 Chapter 3: Differentiation Normal Tangent FIGURE 3. Eample shows how to fin equations for the tangent an normal to the folium of Descartes at (2, ). EXAMPLE Tangent an Normal to the Folium of Descartes Show that the point (2, ) lies on the curve Then fin the tangent an normal to the curve there (Figure 3.). Solution The point (2, ) lies on the curve because its coorinates satisf the equation given for the curve: s2s To fin the slope of the curve at (2, ), we first use implicit ifferentiation to fin a formula for >: A3 B + A3 B - A9B A0B We then evaluate the erivative at s, s2, : ` s2, - 9 a + b 0 s s ` s2, s s Differentiate both sies with respect to. Treat as a prouct an as a function of. Solve for >. The tangent at (2, ) is the line through (2, ) with slope >5: + 5 A - 2B The normal to the curve at (2, ) is the line perpenicular to the tangent there, the line through (2, ) with slope -5>: - 5 s The quaratic formula enables us to solve a secon-egree equation like for in terms of. There is a formula for the three roots of a cubic equation that is like the quaratic formula but much more complicate. If this formula is use to solve the equation for in terms of, then three functions etermine b the equation are ƒs 3 C B C B 6-273

5 3.6 Implicit Differentiation 209 an 2 c-ƒs ; 2-3 a 3-3 C 2 + B Using implicit ifferentiation in Eample was much simpler than calculating > irectl from an of the above formulas. Fining slopes on curves efine b higher-egree equations usuall requires implicit ifferentiation. Derivatives of Higher Orer Implicit ifferentiation can also be use to fin higher erivatives. Here is an eample C 2 - B b. EXAMPLE 5 Fin 2 > 2 if Fining a Secon Derivative Implicitl Solution To start, we ifferentiate both sies of the equation with respect to in orer to fin >. We now appl the Quotient Rule to fin. Finall, we substitute 2 > to epress in terms of an. Rational Powers of Differentiable Functions We know that the rule A B s , when Z 0 a2 b a2 b 2-3, when Z 0 n n n # Treat as a function of. Solve for. hols when n is an integer. Using implicit ifferentiation we can show that it hols when n is an rational number. THEOREM Power Rule for Rational Powers If p>q is a rational number, then p>q is ifferentiable at ever interior point of the omain of sp>q -, an p>q p q sp>q -.

6 20 Chapter 3: Differentiation EXAMPLE 6 (a) (b) (c) A>2 B 2 ->2 A2>3 B 2 3 ->3 A->3 B - 3-7>3 Using the Rational Power Rule 2 for 7 0 for Z 0 for Z 0 Proof of Theorem Let p an q be integers with q 7 0 an suppose that p>q. Then 2 q p q p. Since p an q are integers (for which we alrea have the Power Rule), an assuming that is a ifferentiable function of, we can ifferentiate both sies of the equation with respect to an get q q - pp -. If Z 0, we can ivie both sies of the equation b to solve for >, obtaining q q - pp - q q - p q # p - s p>q q - p q # p - p - p>q p # s p - - s p - p>q q p>q p q sq - p - p q A law of eponents p q # s p>q -, which proves the rule. We will rop the assumption of ifferentiabilit use in the proof of Theorem in Chapter 7, where we prove the Power Rule for an nonzero real eponent. (See Section 7.3.) B combining the result of Theorem with the Chain Rule, we get an etension of the Power Chain Rule to rational powers of u: If p>q is a rational number an u is a ifferentiable function of, then is a ifferentiable function of an u p>q up>q p q us p>q - u, provie that u Z 0 if s p>q 6. This restriction is necessar because 0 might be in the omain of u p>q but not in the omain of u s p>q -, as we see in the net eample.

7 3.6 Implicit Differentiation 2 EXAMPLE 7 Using the Rational Power an Chain Rules (a) (b) function efine on [-, ] $++%++& A - 2 B > A - 2 B -3> (-2) - 2A - 2 B 3> (+++)+++* erivative efine onl on s -, scos ->5-5 scos -6>5 scos Power Chain Rule with u scos -6>5 s -sin ssin scos -6>5 5