finding the slope of tangent line at 4cos[x]sin[y] = 1 at (pi/3,pi/6)?

Question

trying to prepare for my final, thanks, also i have another Q if you can help.

for what value of X in the interval [-pi/2,pi/2] does the graph y = sqrt[3] x + 2cos[x] have a horizonatal tangent?

Answer 1

Finding the slope of tangent line at 4cos[x]sin[y] = 1 at (pi/3,pi/6)

use implicit differentiation:
4cosx(cosy)y' -4sinxsiny =0
y'= 4sinxsiny/4cosxcosy
= sinxsiny/cosxcosy

now when x=pi/3 and y=pi/6
then
the slope = sin(pi/3) sin(pi/6) / cos(pi/3) cos(pi/6)
=1.732(0.577) = 1
so the eq. of the tangent line is:
y-pi/6=x-pi/3

for what value of X in the interval [-pi/2,pi/2] does the graph y = sqrt[3] x + 2cos[x] have a horizonatal tangent?
y'=sqrt(3) -2sinx =0
when sinx=sqrt(3)/2
so x = pi/3 .

Answer 2

Just do some implicit differentiation with the product rule.
Differentiating 4 cos x sin y gives 4 (-sin x)*(sin y) + 4 (cos x)*(cos y * y'). The right hand side differentiates to 0.
So we get -4 sin x sin y + 4 cos x cos y * y' = 0.
Substitute in the point:
-4 * sqrt(3)/2 * 1/2 + 4 * 1/2 * sqrt(3)/2 * y' = 0
So -sqrt(3) + y' * sqrt(3) = 0
y' = 1.

For the second one, just differentiate and make equal to 0:
sqrt(3) - 2sin(x) = 0.
2 sin x = sqrt(3)
sin x = sqrt(3)/2.

That makes x = pi/3.