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)x1+xtanx for which we use quotient rule.
According to quotient rule if Y=(f(x))/(g(x))Y=f(x)g(x)
then (dy)/(dx) = (g(x)(df(x))/(dx)-f(x)(dg(x))/(dx))/((g(x))^2)dydx=g(x)df(x)dx−f(x)dg(x)dx(g(x))2
Here f(x)=xf(x)=x and (df)/(dx)=1dfdx=1 and
g(x)=1+xtanxg(x)=1+xtanx and (dg)/(dx)=tanx+xsec^2xdgdx=tanx+xsec2x
Hence (dy)/(dx) = (1+xtanx-x(tanx+xsec^2x))/(1+xtanx)^2dydx=1+xtanx−x(tanx+xsec2x)(1+xtanx)2
= (1-x^2sec^2x)/(1+xtanx)^21−x2sec2x(1+xtanx)2
and (dy)/(dx)=0dydx=0 when 1-x^2sec^2x=01−x2sec2x=0 or
x^2=cos^2xx2=cos2x or x=+-cosxx=±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)^4d2ydx2=(1+xtanx)2(−2xsec2x−2x2sec2xtanx)−2(1−x2sec2x)(1+xtanx)(tanx+xsec2x)(1+xtanx)4
amd when x=+-cosxx=±cosx
(d^2y)/(dx^2)=(-2(1+-sinx)^2(x^3+tanx))/(1+xtanx)^4d2ydx2=−2(1±sinx)2(x3+tanx)(1+xtanx)4
As (d^2y)/(dx^2)<=0d2ydx2≤0 when x=cosxx=cosx not at x=-cosxx=−cosx
We have a local maxima at x=cosxx=cosx and graph gives the solution as x=0.739x=0.739 as appears from graph below. See the point of intersection of y=xy=x and y=cosxy=cosx is at about x=0.739x=0.739.
graph{(y-x)(y-cosx)=0 [-0.667, 1.833, -0.13, 1.12]}
Graph of x/(1+xtanx)x1+xtanx as it appears around local maxima is given below.
graph{x/(1+xtanx) [-0.0256, 1.2155, -0.0604, 0.5603]}
