Use implicit differentiation to show that a function defined implicitly by sin x +

Question

cos y = 2y has a critical point whenever cos x = 0. Then use the first derivative
test to classify those critical numbers that lies in the interval (−2�, 2�) as relative
maxima or minima.

I KNOW HOW TO DO THE IMPLICIT PART BUT
PLIZ CAN ANY 1 EXPLAIN HOW TO FIND THE CRITTICAL POINTS FOR ME..

Answer 1

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Answer 2

if sinx + cosy = 2y
then cosx - siny dy/dx = 2dy/dx
cosx = 2dy/dx + siny dy/dx
dy/dx(2 - siny)= cosx
dy/dx = cosx/(2-siny)

for critical values, dy/dx = 0
thus cosx =0 , as required.

if cosx = 0 , then x = π/2 or x = 3π/2 or

if x=π/2 , into original ....
sin π/2 + cosy = 2y
1 + cosy = 2y
cosy = 2y - 1 , messy equation to solve (y = appr .83543 )
but it did not say to find the actual point but rather to classify as max/min

so second derivative:
y'' = ( (2-siny)(-sinx) - cosx(-cosy dy/dx)/ (2-siny)^2

if x = π/2
y'' = ( 2 - (# less than 1)(-1) - 0(....) / (positive)
= -/+ = negative
so when x = π/2 , we have a maximum

if x = 3π/2
y'' = (2 - .....)((+1) - 0(....) )/(positive)
= +/+ = +
so when x = 3π/2, we have a minimum

Just realized we are to looks at interval (-2,2)
so for cosx = 0
x = -π/2, π/2

Repeating an evaluation similar to the ones above,
when x = -π/2
y'' = +/+ = +
we would have a minimum

Answer 3

critical point are when the derivative is equal to 0