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?
#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
1. To find y'(x), we can use the implicit differentiation method. Starting with the equation:
xsec(y) - 3y sin(x) = 1
Differentiate both sides of the equation with respect to x:
d/dx(xsec(y) - 3y sin(x)) = d/dx(1)
Using the chain rule, we get:
sec(y) * dy/dx + xsec(y)tan(y) * dy/dx - 3y * cos(x) - 3sin(x) * dy/dx = 0
Grouping the terms with dy/dx together, we have:
dy/dx * (sec(y) + xsec(y)tan(y) - 3sin(x)) = 3y * cos(x) - sec(y)
Finally, we can solve for dy/dx:
dy/dx = (3y * cos(x) - sec(y)) / (sec(y) + xsec(y)tan(y) - 3sin(x))
Comparing this with the answer choices given, we can see that the correct choice is c) (3ycosx - sec y) / (xsecytany - 3sinx).
2. To find the absolute extrema of f(x) = sin(x) + cos(x) on the interval [0, 2π], we can first find the critical points by setting the derivative equal to zero and checking the boundaries of the interval.
Taking the derivative of f(x):
f'(x) = cos(x) - sin(x)
Setting f'(x) = 0:
cos(x) - sin(x) = 0
Rearranging this equation, we get:
sin(x) = cos(x)
Dividing by cos(x) (since cos(x) is not equal to zero for the given interval), we get:
tan(x) = 1
Solving for x, we find that x = π/4 and x = 5π/4 are the critical points.
Next, we evaluate the function f(x) at the critical points and the boundaries of the interval:
f(0) = sin(0) + cos(0) = 0 + 1 = 1
f(π/4) = sin(π/4) + cos(π/4) = sqrt(2)/2 + sqrt(2)/2 = sqrt(2)
f(2π) = sin(2π) + cos(2π) = 0 + 1 = 1
Comparing the values, we can see that the maximum value is sqrt(2) and the minimum value is 1. Therefore, the correct choice is c) max: √2 min: -1.