y = \\frac { x ^ { x } } { x + 1 } \\text { at } x = 1\nenter \\frac { d y } { d x } \\text { here. }

Explanation:

Step1: Find derivative of (x^x)

Let (u = x^x), take natural logarithm: (\ln u=x\ln x). Differentiate both sides: (\frac{1}{u}u^\prime=\ln x + 1), so (u^\prime=x^x(\ln x + 1)).

Step2: Use quotient rule ((\frac{f}{g})^\prime=\frac{f^\prime g - fg^\prime}{g^2})

Here (f = x^x), (f^\prime=x^x(\ln x + 1)), (g=x + 1), (g^\prime = 1). Then (y^\prime=\frac{x^x(\ln x + 1)(x + 1)-x^x\times1}{(x + 1)^2}).

Step3: Substitute (x = 1)

When (x = 1), (x^x=1), (\ln x=0). (y^\prime=\frac{1\times(0 + 1)(1 + 1)-1\times1}{(1 + 1)^2}=\frac{2 - 1}{4}=\frac{1}{4}).

Answer:

(\frac{1}{4})