If 2 y= [ ⁻¹ ( 3 x+ x x- 3 x ) ]^2 x (0, 2 ) then d y d x is equal to

Step-by-step solution

The given expression is 2y = [ ⁻¹ ( 3 x + x x - 3 x ) ]^2 . Divide the numerator and denominator inside the ⁻¹ function by x : 3 + x 1 - 3 x = ( 3 + x ) . Thus, the expression becomes ⁻¹ ( ( 3 + x ) ) . Using the identity ⁻¹ ( ) = 2 - , we have: ⁻¹ ( ( 3 + x ) ) = 2 - ( 3 + x ) = 6 - x . Substituting this back into the equation for 2y : 2y = ( 6 - x )^2 . Differentiating both sides with respect to x : 2 dy dx = 2 ( 6 - x ) (-1) . dy dx = - ( 6 - x ) = x - 6 . Answer: x- 6