Let $y=(x^2+1)^{x\ln x}$.
Since the variable $x$ appears in both the base and the exponent, use logarithmic differentiation.
Take $\ln$ on both sides:
$\Rightarrow \ln y=\ln\left((x^2+1)^{x\ln x}\right)$
Using $\ln(a^b)=b\ln a$, we get:
$\Rightarrow \ln y=x\ln x\cdot \ln(x^2+1)$
Now differentiate both sides with respect to $x$.
Left side:
$\Rightarrow \frac{d}{dx}(\ln y)=\frac{y'}{y}$
Right side needs the product rule. Let $u=x\ln x$ and $v=\ln(x^2+1)$.
Then $u'=\ln x+1$ and $v'=\frac{2x}{x^2+1}$.
So,
$\Rightarrow \frac{y'}{y}=(\ln x+1)\ln(x^2+1)+x\ln x\cdot\frac{2x}{x^2+1}$
Simplify:
$\Rightarrow \frac{y'}{y}=(\ln x+1)\ln(x^2+1)+\frac{2x^2\ln x}{x^2+1}$
Now multiply both sides by $y$:
$\Rightarrow y'=y\left((\ln x+1)\ln(x^2+1)+\frac{2x^2\ln x}{x^2+1}\right)$
Since $y=(x^2+1)^{x\ln x}$,
$\boxed{\therefore f'(x)=(x^2+1)^{x\ln x}\left((\ln x+1)\ln(x^2+1)+\frac{2x^2\ln x}{x^2+1}\right)}$
Answer: A.
