Find the derivative of f(x) = ln(x) / (x2 sin(x)) and y = tan(x) / e^(3x).

We use the product rule for $v$: $\frac{d}{dx}(ab) = a'b + ab'$.
Let $a = x^2$ and $b = \sin(x)$.
$a' = \frac{d}{dx}(x^2) = 2x$.
$b' = \frac{d}{dx}(\sin(x)) = \cos(x)$.
So, $v' = (2x)(\sin(x)) + (x^2)(\cos(x)) = 2x \sin(x) + x^2 \cos(x)$.

Factor out $x$ from the numerator.

The derivative is:

: Find the derivative of $y = \frac{\tan(x)}{e^{3x}}$.

We use the quotient rule: $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$.
Let $u = \tan(x)$ and $v = e^{3x}$.

We use the chain rule for $v$.

Factor out $e^{3x}$ from the numerator.

The derivative is:

: Find the derivative of $f(x) = \frac{2x}{x^2}$.

The derivative is:

: Find the derivative of $y = \frac{\ln(x^2 + 1)}{\sqrt{x}}$.

We use the quotient rule: $\frac{dy}{dx} = \frac{u'v - uv'}{v^2}$.
Let $u = \ln(x^2 + 1)$ and $v = \sqrt{x} = x^{1/2}$.

We use the chain rule for $u$.

We use the power rule for $v$.

To simplify the numerator, find a common denominator, which is $2\sqrt{x}(x^2 + 1)$.

The derivative is:

: Find the derivative of $f(x) = \frac{e^x \ln x}{x^2 + 1}$.

We use the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2}$.
Let $u = e^x \ln x$ and $v = x^2 + 1$.

We use the product rule for $u$: $\frac{d}{dx}(ab) = a'b + ab'$.
Let $a = e^x$ and $b = \ln x$.
$a' = \frac{d}{dx}(e^x) = e^x$.
$b' = \frac{d}{dx}(\ln x) = \frac{1}{x}$.
So, $u' = (e^x)(\ln x) + (e^x)\left(\frac{1}{x}\right) = e^x \ln x + \frac{e^x}{x} = e^x\left(\ln x + \frac{1}{x}\right)$.

Expand the term $(x^2 + 1)\left(\ln x + \frac{1}{x}\right)$:

Substitute this back into the numerator:

Combine terms with $\ln x$:

Note that $x^2 - 2x + 1 = (x-1)^2$.

The derivative is: