Question

1. Find y'(x) when
xsecy - 3y sinx = 1

a) (3ycosx - sec y) / (xsec^2y - 3sinx)
b) (3cosx - sec x) / (xsecytany - 3sinx)
c) (3ycosx - sec y) / (xsecytany - 3sinx)
d) (3ycosx - secytany) / xsec^2y - 3sinx)

This is what I did:

xsecy - 3y sinx = 1
=> sec y + x sec y tan y * y' - 3sin x * y' - 3y cos x = 0
=> y' * (x sec y tan y - 3 sin x) = (3y cos x - sec y)
=> y' = (3y cos x - sec y) / (x sec y tan y - 3 sin x)
So, C. Is that right?

2. Find the absolute extrema of f(x) = sinx+cosx on the interval [0, 2π]

a) max: 2 min: -2
b) max: 1 min: -1
c) max: √2 min: -1
d) max: √2 min: -√2


Is it D?

Answer 1

#1 is correct

#2 is d), I had angles of pi/4 and 5pi/4
resulting in a max of √2 and a min of -√2