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Question

The derivative at of tan11+x21x w.r.t. tan12x1x212x2 at x=0 is

A
14
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B
18
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C
12
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D
1
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Solution

The correct option is B 14
Let u=tan11+x21x & v=tan12x1x212x2

Let x=tanθ

u=tan11+tan2θ1tanθ

=tan1secθ1tanθ

=tan11cosθsinθ

=tan111+2sin2θ22sinθ2cosθ2

=tan12sinθ2sinθ22sinθ2cosθ2

=tan1tanθ2

=θ2=12tan1x=12tan1x=u

Let x=sinθ for v

v=tan12x1±x212x2=tan12sinθ1sin2θ12sin2θ

=tan12sinθcosθcos2θ

=tan1sin2θcos2θ=tan1(tan12θ)

=2θ

=2sin1x

dudx=12×11+x2 dudxx=0=12

dvdx=2×11x2 dvdxx=0=2

dudv=dudxx=0dνdxx=0=122=14

dudv=14 .... option A

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