# Find the maxima for x/(1+xtanx)?

Mar 24, 2017

We have a local maxima at $x = 0.739$

#### Explanation:

A maxima occurs when the derivative of the function equals zero and second derivative is negative. Hence we first find out the derivative of $\frac{x}{1 + x \tan x}$ for which we use quotient rule.

According to quotient rule if $Y = \frac{f \left(x\right)}{g \left(x\right)}$

then $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{g \left(x\right) \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} - f \left(x\right) \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}}}{{\left(g \left(x\right)\right)}^{2}}$

Here $f \left(x\right) = x$ and $\frac{\mathrm{df}}{\mathrm{dx}} = 1$ and

$g \left(x\right) = 1 + x \tan x$ and $\frac{\mathrm{dg}}{\mathrm{dx}} = \tan x + x {\sec}^{2} x$

Hence $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 + x \tan x - x \left(\tan x + x {\sec}^{2} x\right)}{1 + x \tan x} ^ 2$

= $\frac{1 - {x}^{2} {\sec}^{2} x}{1 + x \tan x} ^ 2$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ when $1 - {x}^{2} {\sec}^{2} x = 0$ or

${x}^{2} = {\cos}^{2} x$ or $x = \pm \cos x$

We now workout second derivative
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{{\left(1 + x \tan x\right)}^{2} \left(- 2 x {\sec}^{2} x - 2 {x}^{2} {\sec}^{2} x \tan x\right) - 2 \left(1 - {x}^{2} {\sec}^{2} x\right) \left(1 + x \tan x\right) \left(\tan x + x {\sec}^{2} x\right)}{1 + x \tan x} ^ 4$

amd when $x = \pm \cos x$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{- 2 {\left(1 \pm \sin x\right)}^{2} \left({x}^{3} + \tan x\right)}{1 + x \tan x} ^ 4$

As $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \le 0$ when $x = \cos x$ not at $x = - \cos x$

We have a local maxima at $x = \cos x$ and graph gives the solution as $x = 0.739$ as appears from graph below. See the point of intersection of $y = x$ and $y = \cos x$ is at about $x = 0.739$.

graph{(y-x)(y-cosx)=0 [-0.667, 1.833, -0.13, 1.12]}

Graph of $\frac{x}{1 + x \tan x}$ as it appears around local maxima is given below.

graph{x/(1+xtanx) [-0.0256, 1.2155, -0.0604, 0.5603]}