Question

Show the curve with equation does not have any horizontal tangents. for horizontal tangents make y=0 for y' means y'=0?

`y=cosx/(1+sinx)`

Answer 1

See below.

Explanation:

Start by finding the derivative. This can be done using the quotient rule.

`y' = (-sinx(1 + sinx) - cosx(cosx))/(1 + sinx)^2`

`y' = (-sinx -sin^2x - cos^2x)/(1 + sinx)^2`

`y' = (-sinx - (sin^2x+ cos^2x))/(1 + sinx)^2`

Recall that `sin^2x + cos^2x = 1`.

`y' = (-sinx - 1)/(1 + sinx)^2`

`y' = -(sinx + 1)/(1 + sinx)^2`

`y' = -1/(1 + sinx)`

We are looking for values of `x` where `y' = 0`, meaning the tangent is horizontal.

`0 = -1/(1 + sinx)`

`0 = -1`

Since this is clearly false, there are no solutions, thus, there are no horizontal tangents. The graph of the function will confirm.

As you can see lots of asymptotes, but clearly no horizontal tangents.

Hopefully this helps!