When the tangent is horizontal, dy/dx = 0
Have you taken the derivative?
Set it equal to zero and solve for x
Find the values of x on the curve below at which the tangent is horizontal. Use n as an arbitrary integer. (Enter your answers as a comma-separated list.)
y= (sin(x))/ (2+cos(x))
3 answers
I have, but I keep getting an incorrect final answer.
It would be cosx/-sinx, right?
It would be cosx/-sinx, right?
Oh no!
You can't just differentiate top and bottom of a fraction, you must use the quotient rule
y = (sin(x))/ (2+cos(x))
dy/dx = ( (2+cosx)(cosx) - sinx(-sinx)/(2+cosx)^2
= (2cosx + cos^2 x + sin^2x)/(2+cosx)^2
= (2cosx + 1)/(2 + cosx)^2
= 0
then 2cosx + 1 = 0
cosx = -1/2
x = 2π/3 or x = 4π/3
You can't just differentiate top and bottom of a fraction, you must use the quotient rule
y = (sin(x))/ (2+cos(x))
dy/dx = ( (2+cosx)(cosx) - sinx(-sinx)/(2+cosx)^2
= (2cosx + cos^2 x + sin^2x)/(2+cosx)^2
= (2cosx + 1)/(2 + cosx)^2
= 0
then 2cosx + 1 = 0
cosx = -1/2
x = 2π/3 or x = 4π/3
2 answers