Question

tan^{-1} 2x/1- x^{2} dy/dx ज्ञात कीजिए

Answer 1

Answer:

2 / ( 1 + x² )

Step-by-step explanation:

Given----> y = tan⁻¹ { 2x / ( 1 - x² ) }

To find ----> Derivative of given function.

Solution----> ATQ,

y = tan⁻¹ { 2x / ( 1 - x² ) }

Let, x = tanθ => θ = tan⁻¹ x

=> y = tan⁻¹ { 2tanθ / ( 1 - tan²θ ) }

We know that , tan2θ = 2tanθ / ( 1 - tan²θ ) , applying it we get,

= tan⁻¹ ( tan2θ )

= 2 θ

=> y = 2 tan⁻¹ x

Differentiating with respect to x , we get,

=> dy/dx = 2 d/dx ( tan⁻¹ x )

= 2 { 1 / ( 1 + x² ) }

=> dy/dx = 2 / ( 1 + x² )

Additional information---->

1) d/dx ( Sin⁻¹x ) = 1 / √(1 - x² )

2) d/dx ( Cos¹x ) = - 1 / √(1 - x² )

3) d/dx ( Sec⁻¹x ) = 1 / x √(x² - 1 )

4) d/dx ( Cosec⁻¹x ) = -1/x√(x² - 1)

Answer 2

dy/dx =  2/(1 + x²)  यदि y = Tan⁻¹(2x/(1 - x²))

Step-by-step explanation:

dy/dx ज्ञात कीजिए

y = Tan⁻¹(2x/(1 - x²))

=> Tany  =2x/ (1 - x²)

=> Sec²y (dy/dx)  =   (2x (-1)/(1 - x²)²)(-2x)  + 2/(1 - x²)

=> (1 + Tan²y)(dy/dx)  =  4x²/(1 - x²)²  + 2/(1 - x²)²

=> (1 + Tan²y)(dy/dx)  =  ( 4x²  + 2 - 2x²)/(1 - x²)²

=> (1 + Tan²y)(dy/dx)  =  (2x²  + 2)/(1 - x²)²

=> (1 + Tan²y)(dy/dx)  =  2(x²+ 1)/(1 - x²)²

1 + Tan²y  = 1   +  (2x/ (1 - x²))²

=> 1 + Tan²y  =  (1  + x⁴ - 2x²   + 4x²)/(1 - x²)²

=> 1 + Tan²y  =  (1 + x²)²/(1 - x²)²

(1 + x²)²/(1 - x²)² (dy/dx)  =  2(x²+ 1)/(1 - x²)²

=> dy/dx =  2/(1 + x²)

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