Find the derivatives of (a) f(x) = ln(x3 + 3x) . (

The chain rule states that if $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.
In this case, the outer function is $g(u) = \ln(u)$ and the inner function is $h(x) = x^3 + 3x$.

The derivative of $\ln(u)$ is $\frac{1}{u}$. So, $g'(u) = \frac{1}{u}$.

The derivative of $x^3 + 3x$ is $3x^2 + 3$. So, $h'(x) = 3x^2 + 3$.

Substitute $u = x^3 + 3x$ back into $g'(u)$ and multiply by $h'(x)$.

We can factor out 3 from the numerator and $x$ from the denominator:

The final answer is $\frac{3(x^2 + 1)}{x(x^2 + 3)}$.