Let's solve each part of the problem step by step. Given that $g(x) = x \cdot f(\cos(\pi \cdot x))$:

a) $g(3)$

To find $g(3)$, substitute $x = 3$ into the function:

$g(3) = 3 \cdot f(\cos(\pi \cdot 3))$

Since $\cos(\pi \cdot 3) = \cos(3\pi) = -1$:

$g(3) = 3 \cdot f(-1)$

So, the answer is:

$g(3) = 3f(-1)$

b) $g'(x)$

To find $g'(x)$, we'll use the product rule for differentiation. The product rule states that if $h(x) = u(x) \cdot v(x)$, then $h'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$.

Here, $u(x) = x$ and $v(x) = f(\cos(\pi \cdot x))$.

First, differentiate $u(x) = x$:

$u'(x) = 1$

Next, differentiate $v(x) = f(\cos(\pi \cdot x))$. We need to use the chain rule here. Let $h(x) = \cos(\pi \cdot x)$, then:

$v'(x) = f'(h(x)) \cdot h'(x)$

Differentiate $h(x) = \cos(\pi \cdot x)$ with respect to $x$:

$h'(x) = -\pi \cdot \sin(\pi \cdot x)$

So,

$v'(x) = f'(\cos(\pi \cdot x)) \cdot (-\pi \cdot \sin(\pi \cdot x))$

Now apply the product rule:

$g'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$

$g'(x) = 1 \cdot f(\cos(\pi \cdot x)) + x \cdot \left[ f'(\cos(\pi \cdot x)) \cdot (-\pi \cdot \sin(\pi \cdot x)) \right]$

$g'(x) = f(\cos(\pi \cdot x)) - \pi x \cdot \sin(\pi \cdot x) \cdot f'(\cos(\pi \cdot x))$

c) $g'(x^2)$

For $g'(x^2)$, we substitute $x^2$ for $x$ in the derivative $g'(x)$:

$g'(x^2) = f(\cos(\pi \cdot x^2)) - \pi x^2 \cdot \sin(\pi \cdot x^2) \cdot f'(\cos(\pi \cdot x^2))$

d) $g'(1/2)$

Finally, to find $g'(1/2)$, substitute $x = \frac{1}{2}$ into the derivative $g'(x)$:

$g'\left(\frac{1}{2}\right) = f\left(\cos\left(\pi \cdot \frac{1}{2}\right)\right) - \pi \cdot \frac{1}{2} \cdot \sin\left(\pi \cdot \frac{1}{2}\right) \cdot f'\left(\cos\left(\pi \cdot \frac{1}{2}\right)\right)$

Simplifying further:

$\cos\left(\pi \cdot \frac{1}{2}\right) = 0$

$\sin\left(\pi \cdot \frac{1}{2}\right) = 1$

$g'\left(\frac{1}{2}\right) = f(0) - \frac{\pi}{2} \cdot f'(0)$

Summary of Results:

a) $g(3) = 3f(-1)$

b) $g'(x) = f(\cos(\pi \cdot x)) - \pi x \cdot \sin(\pi \cdot x) \cdot f'(\cos(\pi \cdot x))$

c) $g'(x^2) = f(\cos(\pi \cdot x^2)) - \pi x^2 \cdot \sin(\pi \cdot x^2) \cdot f'(\cos(\pi \cdot x^2))$

d) $g'\left(\frac{1}{2}\right) = f(0) - \frac{\pi}{2} \cdot f'(0)$

Would you like more details on any of these steps, or do you have any questions?

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Related Questions:

1. What is the product rule in differentiation?
2. How does the chain rule apply in the differentiation process?
3. What is the significance of the function $f(x)$ in this problem?
4. How would the results change if $g(x)$ were defined as $x^2 \cdot f(\cos(\pi \cdot x))$?
5. How do you compute $g'(x)$ if $g(x) = x \cdot f(e^x)$?
6. What happens if $g(x)$ is a function of two variables?
7. How can you verify the correctness of $g'(x)$?
8. How does $f(x)$ affect the overall behavior of $g(x)$?

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Tip: When applying the chain rule, always identify the outer and inner functions clearly to avoid mistakes in differentiation.