5.6 The Chain Rule and Implicit Di erentiation

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Alfred Woods
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1 5.6 The Chain Rule and Implici Di ereniaion Objecives Icanhowousehechainruleofindhederivaiveofafuncionwihrespeco aparameer(likeime). Iknowowrieheresulofhederivaiveinermsofheparameer(s)only. I can use implici di ereniaion o deermine he derivaive of a variable wih respec o anoher. Up o his poin, we have focused on derivaives based on space variables x and y. Inpracice, however, hese spacial variables, or independen variables, aredependenonime.therefore,i is useful o know how o calculae he funcion s derivaive wih respec o ime. This requires he chain rule. Le us remind ourselves of how he chain rule works wih wo dimensional funcionals. If we are given he funcion y = f(x), where x is a funcion of ime: x = g(). Then he derivaive of y wih respec o is he derivaive of y wih respec o x muliplied by he derivaive of x wih respec o d = dx dx d The echnique for higher dimensions works similarly. The only di culy is ha we need o consider all he variables dependen on he relevan parameer (ime ). 1. Chain Rule - Case 1: Suppose z = f(x, y) andx = g(),y = h(). Based on he one variable case, we can see ha /d is calculaed as d = f dx x d + f y d In his conex, i is more common o see he following noaion. f The is referred o as a parial, shor for parial derivaive. 2. Chain Rule - Case 2: Paramericequaionsmayhavemorehanonevariable,like and s. This could be ime and arc lengh, for example. In his case, z = f(x, y) andx = g(s, ),y = h(s, ). Then, every derivaive is a parial derivaive. Our formula for his siuaion is 129 of 146

2 z = f x x + f y y Wha we do is ake he derivaive wih respec o each variable, hen ake he derivaive wih respec o he hidden parameric variable. Wrien in our new noion, we have he Similarly if, we waned o look a he derivaive wih respec o s, we d To beer undersand hese echniques, le s look a some examples Examples Example Use he chain rule o find /d for z =cos(4x + y), x =5 4,y = 1 We begin by finding all he necessary derivaives. We pu everyhing in erms of by plugging in he funcions for x and y. Thisgivesushefollowingequaions: 20 = 4sin(4x + y) = +1 4sin(4[54 ]+[1/]) = 4sin 20 = sin(4x + y) = +1 sin(4[54 ]+[1/]) = sin dx d =203 d = 1 2 Now, we jus plug ino he formula d d 130 of 146

3 o ge d = sin sin +1 2 Example Use he chain rule o find /@s for z = x 2 y 2 where x = s cos and y = s sin As we saw in he previous example, hese problems can ge ricky because we need o keep all he informaion organized. Le s walk hrough he soluion of his exercise slowly so we don make any misakes. Our final answer will be in erms of s and only. Firs, we need o find he parials of =2x2 y Nex, we find he parials of @s =sin Now, we plug in wha we found ino our @s = 2xy 2 [cos ]+ 2x 2 y [sin ] We also wan o subsiue and simplify so our answer is sensible. Tha gives = 2(s cos )(s sin ) 2 [cos ]+ 2(s cos ) 2 (s sin ) [sin ] =2s 3 cos 2 sin 2 +2s 3 cos 2 sin 2 We can use he rigonomeric formula ha 2 cos sin =sin2 o = s3 sin of 146

4 Example Use he chain rule o find /@ for z = x 2 y 2 where x = s cos and y = s sin (he same equaion for example 2). The mah works he parial derivaives of he inner = = s cos Now we jus plug in as before and. Wha s @x = 2xy 2 [ s sin ]+ 2x 2 y [s cos ] If we subsiue x and y for heir parameric formulas, we = 2(s cos )(s sin ) 2 [ s sin ]+ 2(s cos ) 2 (s sin ) [s cos ] = 2s 3 cos sin 3 +2s 3 cos 3 sin Again, using 2 cos sin =sin2o simplify as well as cos 2 resul sin 2 =cos2, = 2s3 cos sin 3 +2s 3 cos 3 sin = s 3 (2 sin cos )sin 2 + s 3 (2 sin cos )cos 2 = s 3 sin 2(cos 2 sin 2 ) = s 3 sin 2 cos 2 No all funcions can be nicely wrien in z = form. In hese siuaions, however, you may wan o deermine he parial derivaive of z wih respec o he variables x or y. Toachievehis, we will use a echnique is called implici di ereniaion. Le s firs consider wha happens in he wo-dimensional case. Suppose we have he equaion F (x, y) = of 146

5 Remember ha we can always se a funcion of wo variables o zero by simply moving everyhing o one side. If we waned o find he oal derivaive of he funcion, we would dx =0 Recall ha dx/dx =1. @y dx =0 =) = F x F y In hree dimensions, i works similarly. We assume ha he derivaives of he inpu variables wih respec o each oher is zero (like dx/ =0). Wemayusehesameprocessaboveoge similar resuls. Tha is, we ge he = F x F = F y F z Le s look a some examples Examples Example Find dx for y cos x = x 2 + y 2 Le s begin by wriing he funcion in he correc form. 0=x 2 + y 2 y cos x Now we find F x and F y. F x =2x + y sin x F y =2y cos x Then, dx = 2x + y sin x 2y cos x 133 of 146

6 Example =x 2 y 2 + z 2 2z in. We begin by puing he equaion in he correc form, find he relevan parials, and hen plug 0=x 2 y 2 + z 2 2z 4, F y =2y, F z =2z = 2y 2z 2 = y z 1 Example x 2 y 2 + z 2 2z =4 Le s use he informaion from example 2. All we need o find is F x =2x = 2x 2z 2 = x z 1 Summary of Ideas: Chain Rule and Implici Di ereniaion Someimes x and y are funcions of one or more parameers. We may find he derivaive of a funcion wih respec o ha parameer using he chain rule. The formulas for calculaing such derivaives are d d To calculae a parial derivaive of a variable wih respec o anoher requires implici di = F x, F = F y F z 134 of 146

Chapter 7. Response of First-Order RL and RC Circuits

Chapter 7. Response of First-Order RL and RC Circuits Chaper 7. esponse of Firs-Order L and C Circuis 7.1. The Naural esponse of an L Circui 7.2. The Naural esponse of an C Circui 7.3. The ep esponse of L and C Circuis 7.4. A General oluion for ep and Naural