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2.5  The Chain Rule<br />Goal:  To use the chain rule to differentiate composite functions f(g(x))<br />The Chain Rule:<br />If  y = f(x) is a differentiable function of u, an u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and <br />dydx=dydu∙dudx<br />In function notation:<br />dydxf(gx)=f'(gx)∙g'(x)<br />Another way to look at it –> since u = g(x)<br />dydxf(u)=f'(u)∙u'<br />Think of this as taking the derivative of the outside then the derivative of the inside.<br />Example:<br />y = (x2 + 2x + 5)3<br />u = x2 +2x +5  <br />u’ = 2x +2<br />Rewrite as:<br />y = u3<br />y’ = 3u2u’<br />y’ =3(x2 +2x +5)2(2x +2)<br />Example:<br />y=1x+1<br />y = (x+1)-1/2<br />u = x + 1<br />u’ = 1<br />Rewrite:<br />y = u-1/2<br />y’ =( -1/2)u-3/2u’<br />y'=-12u3u’<br />y'=-12uuu'<br />y'=-u2u2u’<br />y'=-x+12(x+1)2<br /> <br />Example:<br />y = (x3 + 1)2<br />u = x3 + 1<br />u’ = 3x2<br />Rewrite:<br />y = u2<br />y’ = 2u1u’<br />y’ = 2(x3 + 1)( 3x2)<br />y’ = 6x5 + 6x2<br />Example:<br />y = (x2 + 3x)4<br />u = x2 + 3x<br />u’ = 2x + 3<br />Rewrite:<br />y = u4<br />y’ = 4u3u’<br />y’ = 4(x2 + 3x)3( 2x + 3)<br />Example:<br />y=42x+1<br />y = 4(2x+1)-1<br />u = 2x + 1<br />u’ =2<br />Rewrite:<br />y = 4u-1<br />y’ = 4u--2u’<br />y’ = 4u-2(2)<br />y’ = 8u-2<br />y'=8u2<br />y'=-8(2x+1)2<br />Example:<br />y=2(x-1)3<br />y = 2(x - 1)-3<br />u = x - 1<br />u’ =1<br />Rewrite:<br />y = 2u-3<br />y’ = -6u--4u’<br />y’ = -6u-4(1)<br />y'=-6u4<br />y'=-6(x-1)4<br />2.5 Cont. <br />Goal: to use the chain rule with the product and quotient rules<br />Example:<br />y=x2x2+1<br />f(x) = x2g(x) = (x2 + 1)1/2 u = x2+1  &  u’ = 2x<br />f’(x) = 2xg’(x) = ? Use chain rule hereg(x) = u1/2  so  g'x=12uu’ = u'u2u<br />g'x=2xx2+12(x2+1)=xx2+1(x2+1)<br /> y'=x2∙xx2+1(x2+1)+2x∙(x2+1)12<br />y'=x3∙x2+1(x2+1)+2x∙x2+1<br />Example:<br />y      =x+1x-52   u=f(x)g(x)      u’= ? use quotient rule<br />y     =   u2f(x) = x + 1g(x) = x – 5<br />y’     = 2uu’f’(x) = 1g’(x) = 1<br />u'=1x-5-1(x+1)(x-5)2=-6(x-5)2<br />y'=2x+1x-5(-6)(x-5)2<br />y'=-12(x+1)(x-5)3Page 138/44-76 multiples of 4<br />
Chain rule
Chain rule
Chain rule

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Chain rule

  • 1. 2.5 The Chain Rule<br />Goal: To use the chain rule to differentiate composite functions f(g(x))<br />The Chain Rule:<br />If y = f(x) is a differentiable function of u, an u = g(x) is a differentiable function of x, then y = f(g(x)) is a differentiable function of x and <br />dydx=dydu∙dudx<br />In function notation:<br />dydxf(gx)=f'(gx)∙g'(x)<br />Another way to look at it –> since u = g(x)<br />dydxf(u)=f'(u)∙u'<br />Think of this as taking the derivative of the outside then the derivative of the inside.<br />Example:<br />y = (x2 + 2x + 5)3<br />u = x2 +2x +5 <br />u’ = 2x +2<br />Rewrite as:<br />y = u3<br />y’ = 3u2u’<br />y’ =3(x2 +2x +5)2(2x +2)<br />Example:<br />y=1x+1<br />y = (x+1)-1/2<br />u = x + 1<br />u’ = 1<br />Rewrite:<br />y = u-1/2<br />y’ =( -1/2)u-3/2u’<br />y'=-12u3u’<br />y'=-12uuu'<br />y'=-u2u2u’<br />y'=-x+12(x+1)2<br /> <br />Example:<br />y = (x3 + 1)2<br />u = x3 + 1<br />u’ = 3x2<br />Rewrite:<br />y = u2<br />y’ = 2u1u’<br />y’ = 2(x3 + 1)( 3x2)<br />y’ = 6x5 + 6x2<br />Example:<br />y = (x2 + 3x)4<br />u = x2 + 3x<br />u’ = 2x + 3<br />Rewrite:<br />y = u4<br />y’ = 4u3u’<br />y’ = 4(x2 + 3x)3( 2x + 3)<br />Example:<br />y=42x+1<br />y = 4(2x+1)-1<br />u = 2x + 1<br />u’ =2<br />Rewrite:<br />y = 4u-1<br />y’ = 4u--2u’<br />y’ = 4u-2(2)<br />y’ = 8u-2<br />y'=8u2<br />y'=-8(2x+1)2<br />Example:<br />y=2(x-1)3<br />y = 2(x - 1)-3<br />u = x - 1<br />u’ =1<br />Rewrite:<br />y = 2u-3<br />y’ = -6u--4u’<br />y’ = -6u-4(1)<br />y'=-6u4<br />y'=-6(x-1)4<br />2.5 Cont. <br />Goal: to use the chain rule with the product and quotient rules<br />Example:<br />y=x2x2+1<br />f(x) = x2g(x) = (x2 + 1)1/2 u = x2+1 & u’ = 2x<br />f’(x) = 2xg’(x) = ? Use chain rule hereg(x) = u1/2 so g'x=12uu’ = u'u2u<br />g'x=2xx2+12(x2+1)=xx2+1(x2+1)<br /> y'=x2∙xx2+1(x2+1)+2x∙(x2+1)12<br />y'=x3∙x2+1(x2+1)+2x∙x2+1<br />Example:<br />y =x+1x-52 u=f(x)g(x) u’= ? use quotient rule<br />y = u2f(x) = x + 1g(x) = x – 5<br />y’ = 2uu’f’(x) = 1g’(x) = 1<br />u'=1x-5-1(x+1)(x-5)2=-6(x-5)2<br />y'=2x+1x-5(-6)(x-5)2<br />y'=-12(x+1)(x-5)3Page 138/44-76 multiples of 4<br />