A maxima occurs when the derivative of the function equals zero and second derivative is negative. Hence we first find out the derivative of x/(1+xtanx) for which we use quotient rule.
According to quotient rule if Y=(f(x))/(g(x))
then (dy)/(dx) = (g(x)(df(x))/(dx)-f(x)(dg(x))/(dx))/((g(x))^2)
Here f(x)=x and (df)/(dx)=1 and
g(x)=1+xtanx and (dg)/(dx)=tanx+xsec^2x
Hence (dy)/(dx) = (1+xtanx-x(tanx+xsec^2x))/(1+xtanx)^2
= (1-x^2sec^2x)/(1+xtanx)^2
and (dy)/(dx)=0 when 1-x^2sec^2x=0 or
x^2=cos^2x or x=+-cosx
We now workout second derivative
(d^2y)/(dx^2)=((1+xtanx)^2(-2xsec^2x-2x^2sec^2xtanx)-2(1-x^2sec^2x)(1+xtanx)(tanx+xsec^2x))/(1+xtanx)^4
amd when x=+-cosx
(d^2y)/(dx^2)=(-2(1+-sinx)^2(x^3+tanx))/(1+xtanx)^4
As (d^2y)/(dx^2)<=0 when x=cosx not at x=-cosx
We have a local maxima at x=cosx and graph gives the solution as x=0.739 as appears from graph below. See the point of intersection of y=x and y=cosx is at about x=0.739.
graph{(y-x)(y-cosx)=0 [-0.667, 1.833, -0.13, 1.12]}
Graph of x/(1+xtanx) as it appears around local maxima is given below.
graph{x/(1+xtanx) [-0.0256, 1.2155, -0.0604, 0.5603]}
