if g(x) = xf(cos pi * x), find in terms of f

a) g(3)
b) g'(x)
c) g'(x^2)
d) g'(1/2)

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.

Learn how to solve the problem g(x) = x * f(cos(pi * x)), including finding specific values like g(3) and g'(x). Detailed step-by-step solutions using function composition, the chain rule, and the product rule are provided. Suitable for advanced high school students.

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