If f (x)= ⁻¹ ( x^x-x^ -x 2 ) , then the value of f ^ (1) is equal to

Step-by-step solution

Consider the function f (x)= ⁻¹ ( x^x-x^ -x 2 ) . The derivative at x=1 follows from the chain rule and derivative of the argument. Let u(x)= x^x-x^ -x 2 . Then f '(x)= -1 1+u^2 u'(x) . The derivatives of x^x and x^ -x are found via logarithmic differentiation: d dx x^x=x^x(1+ x) and d dx x^ -x =-x^ -x (1+ x) . Thus, the derivative of u(x) is: u'(x)= 1 2 [x^x(1+ x)+x^ -x (1+ x) ]= 1 2 (1+ x)(x^x+x^ -x ) . At x=1 , we compute: u(1)= 1^1-1⁻¹ 2 =0 and u'(1)= 1 2 (1+ 1)(1^1+1⁻¹)= 1 2 (1)(2)=1 . Therefore, f '(1)= -1 1+