For $n\epsilon N$, let $f\left ( x \right )=min\:\left \{ 1-\tan ^{n}x, 1-\sin ^{n}x, 1-x^{n} \right \}$, $x\epsilon \left ( -\cfrac {\pi}{2}, \cfrac {\pi}{2} \right )$. The left hand derivative of $f$ at $x=\cfrac {\pi}{4}$ is

Given, $f\left ( x \right )=min\:\left \{ 1-\tan ^{n}x, 1-\sin ^{n}x, 1-x^{n} \right \}$

For $x\in \left ( 0, \cfrac {\pi}{4} \right )$, we have

$\tan x> x> \sin x$

$\Rightarrow 1-\tan ^{n}x<1 -x^{n}< 1-\sin ^{n}x$

$\Rightarrow f\left ( x \right )=1-\tan ^{n}x$

So, $\displaystyle f\left (\dfrac{ \pi }{4} \right )=min\:\left \{ 0, 1-\left ( \frac{1}{\sqrt{2}} \right )^{n}, 1-\left ( \frac{\pi }{4} \right )^{n} \right \}$

$\Rightarrow f(\dfrac{\pi}{4})=0$

Now, $\displaystyle {f}'\left ( \dfrac{\pi}{4}^- \right )=\lim_{h\rightarrow 0^-}\frac{f\left ( \dfrac{\pi}{4}+h \right )-f\left ( \dfrac{\pi}{4} \right )}{h}$

$\displaystyle =\lim_{h\rightarrow 0^-}\frac{1-\tan ^{n}\left ( \dfrac{\pi}{4}+h \right )}{h}$

$\displaystyle=\lim _{ h\rightarrow 0^{ - } } \dfrac { 1-\left( \frac { 1+\tan { h } }{ 1-\tan { h } } \right) ^{ n } }{ h }$

$\displaystyle =\lim_{h\rightarrow 0^-}\frac{\left ( 1-\tan h \right )^{n}-\left ( 1+\tan h \right )^{n}}{h\left ( 1-\tan h \right )^{n}}$

$=\displaystyle \lim _{ h\rightarrow 0^{ - } } \frac { -2\left[ n\tan { h } +\frac { n(n-1) }{ 2 } \tan ^{ 3 }{ h } +..... \right] }{ h\left( 1-n\tan h+\frac { n(n-1) }{ 2 } \tan ^{ 2 }{ h } +.... \right) }$

As, $\underset { h\rightarrow 0 } \lim { \tan h } =0$

So, using this result , we get

$=\underset { h\rightarrow 0^{ - } } \lim \cfrac { -2n\tan { h } }{ h }$

$=-2n$ $\left (\underset { h\rightarrow 0 } \lim \dfrac{ \tan h } { h }=1\right)$