Question

Question 10: Find dy/dx : y = tan¯¹(3x - x³/1-3x²),-1/√3 < x < √3

Class 12 - Math - Continuity and Differentiability

Answer 1

Refer to the attachment
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Formulas used ;-
•d/dx(tan-¹x) = 1/1+x²
•tan3∅ = (3tan∅ - tan³∅)/1-3tan²∅
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After putting ∅ = tan-¹x
differentiate w.r.t x
we get
dy/dx = 3/1+x²

Answer 2

★ DIFFERENTIATION ★

Given expression for y = tan^-1 ( 3x - x^3 / 1 - 3x² ) for which x has a definite range as ;

-1/√3 < x < √3

y = tan^-1 ( 3x - x^3 / 1 - 3x² )

x = Tan θ

y = tan^-1 ( 3tan θ - tan^3 θ / 1 - 3 tan² θ )

Now , y = 3θ

θ = tan^-1

Furthermore differentiation will result in ,

dy/dx = 3/1 + x²

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