Ostroffmid2au22sol

1. [5 points per part] For each of the following functions, compute f 0 ( x ) (a) f ( x ) = ln(sec( x ) + e x ) (b) f ( x ) = arcsin(3 x 2 ) (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) = Tx In ( cos ( x ) + 2) d ¥ ¥ = Kosh ) + 2) + A j=G* 㱺 ¥ H% ¥ +r ¥

2. [12 points] Consider the following parametric curve on the domain t > 0 : x ( t ) = t 2 - 12 t + 10 ln( t ) y ( t ) = 5 arctan( t ) - t Find all points on the curve where the tangent line is vertical , 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=1 or 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] Let f ( x ) = x p 4 x + 1 . Use the linearization of f at x = 2 to find an approximate solution to the equation x p 4 x + 1 = 5 . 935 . 4. [9 points] Consider the curve defined implicitly by the equation cos( ⇡ x ) + 3 y = x 2 y + 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