How to solve this problem? Please help!?

Question

How to find the derivitive of sinX/secX?
thanks!

Another problem:
How to find all points of the curve of Y=sinXcosX where the tangent line is horizontal?

Thanks

Answer 1

Since sec(x) = 1/cos(x), you can rewrite this as sin(x)cos(x). From there you can take a derivative:

d/dx = sin(x)[-sin(x)] + cos(x)cos(x)
d/dx = -sin^2(x) + cos^2(x)

As for the horizontal tangent line, finding the derivative gives you the tangent, and a horizontal tangent means that you have a 0 slope. So if you set the derivative of your question (which, coincidentally, we just found) to 0, you can find the points:

0 = -sin^2(x) + cos^2(x)
sin^2(x) = cos^2(x)
sin(x) = cos(x)
x = pi/4

So your function sin(x)cos(x) will have a horizontal slope every 45 degrees, or (2n - 1)pi/4 for integer n.