Math 155a_Exam Solution on Equations, Implicit Differentiation, Linear Approximation, Absolute Maxim

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MATH 155A FALL 13SOLUTIONS TO THE PRACTICE MIDTERM 2.Question 1. Findy0.(a)y=1x-15x3.(b)y=tanx1 + cosx.(c)y=x21+πcosx.(d)y=1sin(x-sinx).(e)y= sin2(cos(psin(πx)).Solution. Direct application of the differentiation rules yields:(a)y0=35x5x3-12xx.(b)y0=(1+cosx) sec2x+tanxsinx(1+cosx)2.(c)y0=21 +πx21+π-1cosx-x21+πsinx.(d)y0=-cos(x-sinx)(1-cosx)sin2(x-sinx).(e)y0=-πsin(cossin(πx)) cos(cossin(πx)) sinsin(πx) cos(πx)sin(πx).Question 2. Find an equation for the tangent line and normal line to the curve at the givenpoint.(a)y= (1 + 2x)2,(1,9).(b)y=xx+ 1,(4,0.4).(c)x2+ 2xy-y2+x= 2,(1,2).Solution.
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2MATH 155A FALL 13The slope of the normal line is-112, hence the normal equation isy-9 =-112(x-1).(b) Similar to (a); the two equations arey-0.4 =-0.03(x-4) andy-0.4 =1003(x-4).(c) Use implicit differentiation to gety0=-2x-2y-12x-2y.

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