MATH 1550 : Test2 Solutions
Document Summary
In questions 1 5, nd f (x) for the given function f . You do not need to simplify your answers: f (x) = 3. Solution: f (x) = sec x x2+5. 6x1/2 f (x) = (x2 + 5) sec x tan x 2x sec x (x2 + 5)2 f (x) = 2 x2(4 x2) 3/2 f (x) = (ln 4) csc2 x4cot x: f (x) = (ex. Solution. f (x) = ex tan 1 x + ex. 1 + x2: find an equation of the tangent line to the curve y = e3x point (0, 1). We have y = 3e3x tangent line is. Cos x and y (0) = 2, so an equation of the y 1 = 2x: use implicit di erentiation to nd y if 2x2 + y2 = 5. Do simplify your answer as much as possible. We have 4x + 2yy = 0, so y = 2 x y xy y .
