Thread: 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

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

Use the Quotient Rule to take the derivative...







.


Set the derivative equal to 0 and solve for 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.

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

Image taken from http://webgraphing.com/graphing_advanced.jsp



You can see that nowhere on the graph is there a horizontal tangent.