Question

y = \frac { x ^ { x } } { x + 1 } \text { at } x = 1
enter \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:

\(\frac{1}{4}\)