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Find dy/dx in the following: ЁЭСж = tanтИТ1 тБб(3тБвЁЭСетИТЁЭСе3/1тИТ3тБвЁЭСе2),тИТ1тИЪ3 <ЁЭСе <1тИЪ3 - Mathematics

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Find `bb(dy/dx)` in the following:

`y = tan^(-1) ((3x -x^3)/(1 - 3x^2)), - 1/sqrt3 < x < 1/sqrt3`

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y = `tan^-1 ((3x - x^3)/(1 - 3x^2))`

Putting x = tan θ,

∴ y = `tan^-1 ((3 tan theta - tan^3 theta)/(1 - 3 tan^2 theta))`

= tan−1 (tan 3 θ)

= 3 θ 

= 3 tan−1 x  ...[because θ = tan−1 x]

On differentiating with respect to x,

`dy/dx = 3 d/dx tan^-1 x`

`dy/dx = 3 xx 1/(1 + x^2)`

`dy/dx = 3/(1 + x^2)`

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рдкрд╛рда 5: Continuity and Differentiability - Exercise 5.3 [рдкреГрд╖реНрда резремреп]

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рдПрдирд╕реАрдИрдЖрд░рдЯреА Mathematics Part 1 and 2 [English] Class 12
рдкрд╛рда 5 Continuity and Differentiability
Exercise 5.3 | Q 10 | рдкреГрд╖реНрда резремреп

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