find the x-cordinates on the curve y = x tan x + 2 sin x / t

[kimmy_koo51]

find the x-cordinates on the curve y = x tan x + 2 sin x / tan x at which the tangent line is horizontal

So far I have:

y' = (tanx + x sec^2 x + 2 cos x)(tan x) - (x tan x + 2 sin x)(sec^2 x) / tan^2 x

I know that when the tangent line is horizontal the slope is 0 which means that you must set y' = 0 and solve for x. I have no idea how to solve for x in this instance.

 

[o_O]

Try simplifying y first before you differentiate it:

\(\displaystyle 2\frac{sinx}{tanx} = 2\frac{\: sinx \:}{\frac{sinx}{cosx}} = \: ?\)

Then set y' to 0 and it should be easier to solve.

 

[Dr. Flim-Flam]

By graphing this function, one sees only one horizontal line (maybe).

Hence f(x) = xtan(x) + 2sin(x)/tan(x) = xtan(x) + 2cos(x).

dy/dx = tan(x) + xsec²(x) - 2sin(x). Letting tan(x) + xsec²(x) - 2sin(x) = 0,

by inspection we see x = 0. Ergo f(0) = 2, therefore y =2 is the horizontal line.

By using the technological advances that man has given us avoids a lot of grunt work

as sloving this derivative algebraically looks like a lot of ingeniuty is needed.

 

[galactus]

Is that \(\displaystyle \frac{xtan(x)+2sin(x)}{tan(x)}\)?.

If so, then it boils down to something easy.

\(\displaystyle 1-2sin(x)\)

But, if it is \(\displaystyle xtan(x)+\frac{2sin(x)}{tan(x)}\), then that's another matter.