# Thread: derivative of trig function and horizontal tangent lines

1. ## derivative of trig function and horizontal tangent lines

I think I am going to start helping people in algebra or something to earn my keep here. Anyways, here is the one I am workin on.

Find where graph of y= cosx/ (1+sinx) has horizontal tangent lines on the interval (0, pi/2)

I have simplified the derivative down to:

(-sinx-cos^2x)/ (1+ sinx) and I know that i need to find where x = 0 i am just at a loss as to how to find it. Any help is greatly appreciated, as always.

Thanks

Jordan

2. Originally Posted by jwbehm
I think I am going to start helping people in algebra or something to earn my keep here. Anyways, here is the one I am workin on.

Find where graph of y= cosx/ (1+sinx) has horizontal tangent lines on the interval (0, pi/2)

I have simplified the derivative down to:

(-sinx-cos^2x)/ (1+ sinx) and I know that i need to find where x = 0 i am just at a loss as to how to find it. Any help is greatly appreciated, as always.

Thanks

Jordan
To have horizontal tangent lines, the derivative is 0.

So $y = \frac{\cos{x}}{1 + \sin{x}}$

Use the Quotient Rule to take the derivative...

$\frac{dy}{dx} = \frac{(1 + \sin{x})(-\sin{x}) - \cos{x}(\cos{x})}{(1 + \sin{x})^2}$

$= \frac{-\sin{x} - \sin^2{x} - \cos^2{x}}{(1 + \sin{x})^2}$

$= \frac{-\sin{x} - 1}{(1 + \sin{x})^2}$

$= -\frac{1}{1 + \sin{x}}$.

Set the derivative equal to 0 and solve for x...

$0 = -\frac{1}{1 + \sin{x}}$.

I think you can see here that x does not have a solution. So there does not exist any x that will give the original function horizontal tangent lines.

3. Originally Posted by jwbehm
I think I am going to start helping people in algebra or something to earn my keep here. Anyways, here is the one I am workin on.

Find where graph of y= cosx/ (1+sinx) has horizontal tangent lines on the interval (0, pi/2)

I have simplified the derivative down to:

(-sinx-cos^2x)/ (1+ sinx) and I know that i need to find where x = 0 i am just at a loss as to how to find it. Any help is greatly appreciated, as always.

Thanks

Jordan