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Find dy/dx in the following: ๐‘ฆ = tanโˆ’1 โก(3โข๐‘ฅโˆ’๐‘ฅ3/1โˆ’3โข๐‘ฅ2),โˆ’1โˆš3 <๐‘ฅ <1โˆš3 - Mathematics

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Question

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

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

Sum
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Solution

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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Chapter 5: Continuity and Differentiability - Exercise 5.3 [Page 169]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 5 Continuity and Differentiability
Exercise 5.3 | Q 10 | Page 169

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