We are given the function $f(x) = (x+1)^5(x-1)^3$, and we are tasked with finding $f'(x)$, the derivative of this function.

Step-by-step process:

This function is a product of two terms:

1. $(x+1)^5$
2. $(x-1)^3$

We will apply the product rule for differentiation, which states: $(fg)' = f'g + fg'$ Let $u(x) = (x+1)^5$ and $v(x) = (x-1)^3$.

Differentiate $u(x) = (x+1)^5$:

Using the power rule: $u'(x) = 5(x+1)^4$

Differentiate $v(x) = (x-1)^3$:

Using the power rule again: $v'(x) = 3(x-1)^2$

Now, applying the product rule: $f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$ Substitute the values we have: $f'(x) = 5(x+1)^4(x-1)^3 + (x+1)^5 \cdot 3(x-1)^2$ Factor out the common terms: $f'(x) = (x+1)^4(x-1)^2 \left[ 5(x-1) + 3(x+1) \right]$

Simplify inside the brackets: $f'(x) = (x+1)^4(x-1)^2 \left[ 5x - 5 + 3x + 3 \right]$ $f'(x) = (x+1)^4(x-1)^2 \left[ 8x - 2 \right]$

Finally: $f'(x) = 2(x+1)^4(x-1)^2(4x-1)$

Thus, the correct answer is: $\boxed{2(x+1)^4(x-1)^2(4x-1)}$

Let me know if you'd like further explanation or have any questions!

Related questions:

1. How is the product rule applied to more complex functions?
2. What is the power rule and how is it used in differentiation?
3. Can the product rule be applied to more than two functions?
4. How would the chain rule be used if the powers were functions of $x$?
5. How does the quotient rule differ from the product rule?

Tip:

When applying the product rule, always identify the two parts of the product clearly and take their derivatives separately before combining using the rule!