Differential Calculus> let f(x)=min{1-tan^n x,1-sin^n x,1-x^n } ...

let f(x)=min{1-tan^n x,1-sin^n x,1-x^n } xbelongs to(-pi/2,pi/2) where n belong to N.Then the left hand derivative of the f(x)atx=pi/4 is :
A.-2n
b.-2n+1
c.-n(pi/4)^n-1
D.n(pi/4)^ n-1

rajat , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Ans:

$f(x) = min(1-sin^{n}x ,1 -x^{n},1-tan^{n}x)$

$sinx < x < tanx$

$sin^{n}x < x^{n} < tan^{n}x$

$-sin^{n}x > -x^{n} > -tan^{n}x$

$1-sin^{n}x >1 -x^{n} > 1-tan^{n}x$

$LHD = \lim_{h\rightarrow 0}\frac{((1-tan^{n}(\frac{\pi }{4}-h))-(1-1))}{-h}$

$LHD = \lim_{h\rightarrow 0}\frac{(1-tan^{n}(\frac{\pi }{4}-h)}{-h}$

It is zero by zero indeterminent form. Apply L’Hospital Rule,

$LHD = \lim_{h\rightarrow 0}\frac{(ntan^{n-1}(\frac{\pi }{4}-h).sec^{2}(\frac{\pi }{4}-h)}{-1} = -2n$

Thanks & Regards

Jitender Singh

IIT Delhi

askIITians Faculty

Last Activity: 11 Years ago

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Differential Calculus> let f(x)=min{1-tan^n x,1-sin^n x,1-x^n } ...

let f(x)=min{1-tan^n x,1-sin^n x,1-x^n } xbelongs to(-pi/2,pi/2) where n belong to N.Then the left hand derivative of the f(x)atx=pi/4 is :A.-2nb.-2n+1c.-n(pi/4)^n-1D.n(pi/4)^ n-1

rajat , 11 Years ago

Jitender Singh

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