MHT CET202520 Apr 2025Morning ShiftMathematicsDifferentiationActual
If y =x^x+x^ 1 x , then dy d x is equal to
Options
- Ax^x(1+ x)+x^ 1 x 1 x^2 (1- x)
- B(x^x+x^ 1 x ) [1+ x+ 1 x^2 (1- x) ]
- C(x^x+x^ 1 x ) [(1+ x)- 1 x^2 (1- x) ]
- Dx^x(1+ x)-x^ 1 x 1 x^2 (1- x)
Correct answer
A. x^x(1+ x)+x^ 1 x 1 x^2 (1- x)
Step-by-step solution
Differentiate y = x^x + x^ 1/x by handling each term separately. Let u = x^x and v = x^ 1/x , so dy dx = du dx + dv dx . For u = x^x , apply logarithmic differentiation: u = x x . Differentiate both sides: 1 u du dx = x + 1 , so du dx = x^x(1 + x) . For v = x^ 1/x , use logarithmic differentiation: v = 1 x x . Differentiate: 1 v dv dx = - x x^2 + 1 x^2 = 1 - x x^2 , so dv dx = x^ 1/x 1 - x x^2 . Combine the results: dy dx = x^x(1 + x) + x^ 1/x 1 - x x^2 . This matches option A.