《微积分隐分化 en》由会员分享，可在线阅读，更多相关《微积分隐分化 en（11页珍藏版）》请在金锄头文库上搜索。
1、Section 2.6 Implicit Differentiation2010 Kiryl TsishchankaImplicit DifferentiationSome functions can be described by expressing one variable explicitly in terms of another variable for example,y = x2,y =1 x 1 + x3,y = tan2xor, in general, y = f(x). Some functions, however, are defined implicitly by
2、a relation between x and y such thatx2+ y2= 1,x3+ y3= 4xy,(x2+ y2 2x)2= 4(x2+ y2)In some cases it is possible to solve such an equation for x or for y, but sometimes it is impossible.One of the main goals of Section 2.6 is to show how to find derivatives of implicitly defined functions.EXAMPLE: Find
3、 an equation of the tangent line to y = x2at the point (2,4).Solution: The graph of the curve y = x2is a parabola. Clearly,y0= (x2)0= 2xTo find an equation of the tangent line to y = x2at the point (2,4) we note that m = y0(2) = 2 2 = 4therefore the equation isy 4 = 4(x 2)=y = 4x 4EXAMPLE: Find equa
4、tions of the tangent lines to x = y2at the points (4,2) and (4,2).1Section 2.6 Implicit Differentiation2010 Kiryl TsishchankaEXAMPLE: Find equations of the tangent lines to x = y2at the points (4,2) and (4,2).Solution 1: The graph of the curve x = y2is a parabola. We havex = y2=y = xif y 0xif y 0Cle
5、arly, if y =x, theny0= (x1/2)0=1 2x1/21=1 2x1/2=1 2x(1)Similarly, if y = x, theny0= (x1/2)0= 12x1/21= 12x1/2= 1 2x(2)To find an equation of the tangent line to y2= x at the point (4,2) we note that by (1) we have m = y0(4) =1 24=1 4 therefore the equation isy 2 =1 4(x 4)=y =1 4x + 1Similarly, to fin
6、d an equation of the tangent line to y2= x at the point (4,2) we note that by (2) we havem = y0(4) = 1 24= 14 therefore the equation isy (2) = 14(x 4)=y = 14x 1Solution 2: To find equations of the tangent lines to x = y2at the points (4,2) and (4,2) wefirst finddy dxby differentiating both sides of
7、x = y2:x = y2=x0= (y2)0=1 = 2y y0=y0=1 2yIt follows thatm = y0(4) = 1 2 2if y = 21 2 (2)if y = 2therefore the equations arey 2 =1 4(x 4)=y =1 4x + 1at(4,2)and y (2) = 14(x 4)=y = 14x 1at(4,2)2Section 2.6 Implicit Differentiation2010 Kiryl TsishchankaEXAMPLE: If x2+ y2= 5, finddy dx. Then find an equ
8、ation of the tangent line to x2+ y2= 5at the point (2,1).Solution: The graph of the curve x2+ y2= 5 is a circle:221y0 1-1-20x-1-2To finddy dxwe differentiate both sides:x2+ y2= 5=(x2+ y2)0= 50=(x2)0+ (y2)0= 50=2x x0+ 2y y0= 0hence2x 1 + 2y y0= 0=2x + 2y y0= 0=2y y0= 2x=y0= 2x2y= xyTo find an equatio
9、n of the tangent line to x2+ y2= 5 at the point (2,1) we note thatm = y0(2) = 21= 2therefore the equation isy 1 = 2 (x 2)=y = 2x + 5EXAMPLE: If 2x2+3y2= 5, finddy dx. Then find an equation of the tangent line to 2x2+3y2= 5at the point (1,1).3Section 2.6 Implicit Differentiation2010 Kiryl Tsishchanka
10、EXAMPLE: If 2x2+3y2= 5, finddy dx. Then find an equation of the tangent line to 2x2+3y2= 5at the point (1,1).Solution: The graph of the curve 2x2+ 3y2= 5 is an ellipse:x10.5-10y0 -0.5-0.5-1.51.510.5-1To finddy dxwe differentiate both sides:2x2+ 3y2= 5=(2x2+ 3y2)0= 50=2(x2)0+ 3(y2)0= 50=4x + 6y y0= 0
11、hence y0= 4x6y= 2x3yTo find an equation of the tangent line to 2x2+ 3y2= 5 at the point (1,1) we note thatm = y0(1) = 2 13 1= 23therefore the equation isy 1 = 23(x 1)=y = 23x +5 3EXAMPLE: If x3+y3= 4xy, finddy dx. Then find an equation of the tangent line to x3+y3= 4xyat the point (2,2). At what poi
12、nt in the first quadrant is the tangent line horizontal?4Section 2.6 Implicit Differentiation2010 Kiryl TsishchankaEXAMPLE: If x3+y3= 4xy, finddy dx. Then find an equation of the tangent line to x3+y3= 4xyat the point (2,2). At what point in the first quadrant is the tangent line horizontal?Solution
13、: The graph of the curve x3+ y3= 4xy is the folium of Descartes:-3-1-2x1y-1-20-31202(a) To finddy dxwe differentiate both sides:x3+ y3= 4xy=(x3+ y3)0= (4xy)0=(x3)0+ (y3)0= (4xy)0Since (4xy)0= 4(xy)0= 4(x0y + xy0) = 4(1 y + xy0) = 4(y + xy0), this gives us3x2+ 3y2 y0= 4(y + xy0)=3x2+ 3y2 y0= 4y + 4xy
14、0=3y2y0 4xy0= 4y 3x2hencey0(3y2 4x) = 4y 3x2=y0=4y 3x2 3y2 4x(b) To find an equation of the tangent line to x3+ y3= 4xy at the point (2,2) we note thatm = y0(2) =4 2 3 22 3 22 4 2= 1therefore the equation isy 2 = 1 (x 2)=y = x + 4-22-4-40-2y4402x5Section 2.6 Implicit Differentiation2010 Kiryl Tsishc
15、hanka(c) The tangent line is horizontal if y0= 0. Using the expression for y0from part (a), we see that y0= 0 when 4y 3x2= 0 (provided that 3y2 4x 6= 0). We have4y3x2= 0=y =3 4x2x3+y3=4xy=x3+(34x2)3 = 4x(34x2) =x3+2764x6= 3x3which gives 27 64x6= 2x3Dividing both sides by 2, we get27 128x6= x3. Since x 6= 0 in the first quadrant, we can divideboth sides by x3which implies27 128x3= 1=x3=128 27hencex =3128 27= 3128327=364 2327=36432327=4323=4 332 1.6798947Plugging in this into y =3 4x2, we gety =3 4(4332)2=4 334 2.1165347Thus the tangent line is horizontal at(4332,4 334), which is
