MHT CET202618 April 2026Evening ShiftMathematicsDifferentiationActual
If f'(x) = ^2 x and y = f ( 2x-1 x^2+1 ) , then dy dx at x = 1 is
Options
- A1 4 ( 1 2 )
- B1 4 ^2 ( 1 2 )
- C^2 ( 1 4 )
- D1 2 ^2 ( 1 2 )
Correct answer
D. 1 2 ^2 ( 1 2 )
Step-by-step solution
Given y = f ( 2x-1 x^2+1 ) Differentiating with respect to x using the chain rule: dy dx = f' ( 2x-1 x^2+1 ) d dx ( 2x-1 x^2+1 ) dy dx = f' ( 2x-1 x^2+1 ) [ (x^2+1)(2) - (2x-1)(2x) (x^2+1)^2 ] Substituting x = 1 : . dy dx |_ x=1 = f' ( 2(1)-1 1^2+1 ) [ (1^2+1)(2) - (2(1)-1)(2(1)) (1^2+1)^2 ] . dy dx |_ x=1 = f' ( 1 2 ) [ 4 - 2 4 ] . dy dx |_ x=1 = f' ( 1 2 ) 1 2 Since f'(x) = ^2 x , we have f' ( 1 2 ) = ^2 ( 1 2 ) . dy dx |_ x=1 = 1 2 ^2 ( 1 2 ) Answer: 1 2 ^2 ( 1 2 )