Show that $\frac{\tan x}x$ with $x\in (0, \frac\pi 2)$ is a strictly increasing function

Question

Show that $\dfrac{\tan(x)}{x}$, $x\in \left(0, \dfrac{\pi}{2}\right)$ is a strictly increasing function.

$$f(x)=\frac{\tan(x)}{x},\quad f'(x)={\frac {x\tan^2(x)-\tan (x) +x}{{x}^{2}}}.$$ How to conclude?

Answer 1

Hint:

$$\frac{x\sec^2x-\tan x}{x^2}\ge0\iff \sec^2x\ge\frac{\tan x}x\iff\frac{2\sin2x}{x}>0\;,\;\;\forall\,x\in\left(0,\,\frac\pi2\right)$$

Answer 2

Another approach:

It is enough to prove the log is increasing (in other words, consider the logarithmic derivative. $$\frac{f'(x)}{f(x)}=\frac{\tan' x}{\tan x}-\frac 1x=\frac1{\sin x\cos x}-\frac 1x=\frac2{\sin 2x}-\frac1 x=\frac{2x-\sin 2x}{x\sin 2x}>0,$$ since the denominator is positive on $(0,\pi/2)$ and $\sin u0$ ($\sin u$ is a convex function on $[0,\pi]$.

Answer 3

Remember, a function is $increasing$ if the firstderivative is positive on that interval.

Our derivative is: $$\dfrac{x\sec^2\left(x\right)-\tan\left(x\right)}{x^2}$$

Compute when this derivative is greater then $0$.

Answer 4

$$y=\frac{\tan{x}}{x} \Rightarrow y'=\frac{x\sec^2{x}-\tan{x}}{x^2}.$$ For an increasing function, $y'>0$, hence$$x\frac{1}{\cos^2{x}}>\frac{\sin{x}}{\cos{x}}\Rightarrow x > \sin{x}\cos{x}\Rightarrow 2x> \sin{2x},$$ in other words $\theta>\sin{\theta}$. This is true for all $\theta \in (0, +\infty)$.