Answer
$$
h(x)=x^{x}
$$
Use the ideas from Exercise 53 to find the derivative of the given function.
$$
\frac{d}{d x} h(x)=u(x)^{v(x)}\left[\frac{v(x) u^{\prime}(x)}{u(x)}+(\ln u(x)) v^{\prime}(x)\right]
$$
where
$$
h(x)=x^{x} ,\quad u(x)=x, \quad v(x)=x ,\quad u^{\prime}(x)=1, \quad v^{\prime}(x)=1
$$
So,
$$
\begin{aligned} h^{\prime}(x) &=\left(x^{2}+1\right)^{5 x}\left[\frac{5 x(2 x)}{x^{2}+1}+\ln \left(x^{2}+1\right) \cdot(5)\right] \\ &=\left(x^{2}+1\right)^{5 x}\left[\frac{10 x^{2}}{x^{2}+1}+5 \ln \left(x^{2}+1\right)\right]. \end{aligned}
$$
Work Step by Step
$$
h(x)=x^{x}
$$
Use the ideas from Exercise 53 to find the derivative of the given function.
$$
\frac{d}{d x} h(x)=u(x)^{v(x)}\left[\frac{v(x) u^{\prime}(x)}{u(x)}+(\ln u(x)) v^{\prime}(x)\right]
$$
where
$$
h(x)=x^{x} ,\quad u(x)=x, \quad v(x)=x ,\quad u^{\prime}(x)=1, \quad v^{\prime}(x)=1
$$
So,
$$
\begin{aligned} h^{\prime}(x) &=\left(x^{2}+1\right)^{5 x}\left[\frac{5 x(2 x)}{x^{2}+1}+\ln \left(x^{2}+1\right) \cdot(5)\right] \\ &=\left(x^{2}+1\right)^{5 x}\left[\frac{10 x^{2}}{x^{2}+1}+5 \ln \left(x^{2}+1\right)\right]. \end{aligned}
$$
