1.[5 points per part]For each of the following functions, computef0(x) (a)f(x) = ln(sec(x) +ex) (b)f(x) = arcsin(3x2) (c)f(x) = (cos(x) + 2) p x f ' (x ) =se¥¥,← de- it of semtex fat G- = , y = ( cos a) + 2) ☒ In(y )= In (404×1+2) F) = TxIn( cos(x ) + 2) d ¥ ¥ = Kosh ) + 2) + A j=G*㱺¥H%¥+r¥

2.[12 points]Consider the following parametric curve on the domaint >0: x(t) =t2 -12t+ 10 ln(t)y(t) = 5 arctan(t)-t Find all points on the curve where the tangent line isvertical, and all points where it is horizontal. (Specify which is which.) Write your answers in exact form. You do not need to simplify. T T ✗ ' 1+7-0, y' G) =/ 0 y '(e) =p ✗ '( t)-10 ✗ ' (e) =2t-12T¥ __ 0 y' Lt)= ¥2 -1=0 21-2-121-+10=0 5 㱺 =/ t' - 6++5=0 5=1+1-2 (t - 1)( t -53=0 + 2=4 7=1or 5+ = 2 ( not -2 b/c of domain ) Horizontal at 1=2 ⊥ f-201-loh.la?5arctan(2)-2)@ Vertical at c-=/& c- = 5 a \ fll.5-t-DIC-35tlol.IS?5arctan(5)-5)fy

3.[9 points]Letf(x) =xp 4x+ 1. Use the linearization offatx= 2to find an approximate solution to the equation xp 4x+ 1 = 5.935. 4.[9 points]Consider the curve defined implicitly by the equationcos(⇡x) + 3y=x2y+ 6. Find the equation of the line tangent to this curve at the point(1,2). f ' (x ) = 45×+1 + 24T¥ f ' (2)=3 + 8-= B- f (2) = 6 L G) = 6t¥(x-2)← 5. 935= 6 t B- ( x - 2) -0.065= ¥ ( x - 2) - 0. 015 = ✗ - 2 ×=1. - ÷ . - Tis infix ) +3% (3)dd¥ = 2 ✗y + ✗ ¥1 5dL (3) dd¥ - ✗ 28×1=2 xy+ its infix ) * = " " ¥ y;¥(x-D

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