Question

Let $f\left(x\right)=min\{1-{\tan }^{n}x,1-{\sin }^{n}x,1-x^n\}\forall xϵ(-\frac{\pi }{2},\frac{\pi }{2})$, where $nϵN$.Then the left hand derivative of $f\left(x\right)$ at $x=\frac{\pi }{4}$is

• A. $-2n$
• B. $-2(n+1)$
• C. $-n{\left(\frac{\pi }{4}\right)}^{n-1}$
• D. $n{\left(\frac{\pi }{4}\right)}^{n-1}$

Answer 1

The correct option is A $-2n$

For $xϵ(0,\frac{\pi }{4}),\tan x>x>\sin x\Rightarrow 1-{\tan }^{n}x<1-x^n<1-{\sin }^{n}x$

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

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

$f^{\prime }\left(\frac{\pi }{4}\right)=\underset{h\to 0-}{lim}\frac{f(\frac{\pi }{4}+h)-f\left(\frac{\pi }{4}\right)}{h}$

$=\underset{h\to 0-}{lim}\frac{1-{\tan }^n(\frac{\pi }{4}+h)}{h}=\underset{h\to 0-}{lim}\frac{{(1-\tan h)}^n-{(1+\tan h)}^n}{h{(1-\tan h)}^n}$

$\underset{h\to 0-}{lim}\frac{-2\tan h{(1-\tan h)}^{n-1}+{(1-\tan h)}^{n-2}+(1+\tan h)+...+{(1+\tan h)}^{n-1}}{h}$

$=-2[1+1+...+1]ntimes=-2n$