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Question

If y=cos11x21+x2then dydxat x=2 will be

A
25
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B
25
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C
Does not exist
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D
None of the above
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Solution

The correct option is B 25
Let’s first do it by putting x=tan θ or θ=tan1x Doing this we get y=cos1(1tan2θ)1+tan2θ .
But, 1tan2θ1+tan2θ=cos2θ (It’s a standard trigonometric identity).
i.e. y=cos1cos2θNow cos1cosx=xi.e.cos1 cos2θ=2θSo dydx=21+x2 [Usingddx(tan1x)=11+x2]
The solution looks correct but it’s not.
Look θ=tan1x implies that π2<θ<π2
Also cos1cosx=x only when 0xπ
But observe that in the above solution we have used cos1cosx=x for the entire domain which is not correct. Let’s correct it then.
cos1cos2θ=2θ only when θ2θπ or 0θπ2
So for 0θπ2, the domain of tan1x i.e. x will be 0 to
i.e for 0θπ2, 0x<
So the above result is correct only for positive x and is not correct for negative x.
Also cos1cos2θ=2θ ifπ2θ<0 orπ2θ<0i.e.,<x<0 (Check the domain of tan1x forπ2θ<0)
So for negative x
cos1cos2θ=2θ=2tan1xi.e.for x<0 y=2tan1xSo dydx=21+x2for x<021+x2for x0
To get its value at -2, we will put x = -2 in 21+x2 because this is the derivate for x < 0.
Sodydxx=2=21+4=25
Using chain Rule.
y=cos11x21+x2dydx=11(1x21+x2)2×2x(1+x2)(1x2)2x(1+x2)2
[Using chain rule and standard formula for derivative of cos1x]
=(1+x2)(1+x2)2(1+x2)2×2x.2(1+x2)2=4x4x2(1+x2)=2x|x|(1+x2)=4x4x2(1+x2)=2x|x|(1+x2) (Please note that 4x2=2|x|and not 2x)
So dydx=⎪ ⎪ ⎪⎪ ⎪ ⎪21+x2for x<0does not exist at x=021+x2when x>0⎪ ⎪ ⎪⎪ ⎪ ⎪
As |x|=x for x>0and x for x<0

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