Let \( f(x)=x^{3} \cdot\left(\frac{x+4}{x+6}\right)^{7} \) \( f^{\prime}(x)= \)

To find the derivative \( f'(x) \) of the function \[ f(x) = x^3 \left( \frac{x+4}{x+6} \right)^7, \] we will use the product rule and the chain rule. The product rule states that if \( u(x) \) and \( v(x) \) are functions of \( x \), then \[ (uv)' = u'v + uv'. \] In our case, we can let \[ u(x) = x^3 \quad \text{and} \quad v(x) = \left( \frac{x+4}{x+6} \right)^7. \] First, we need to find \( u'(x) \): \[ u'(x) = 3x^2. \] Next, we need to find \( v'(x) \). To do this, we will use the chain rule. Let \[ g(x) = \frac{x+4}{x+6}. \] Then \[ v(x) = g(x)^7. \] Using the chain rule, we have \[ v'(x) = 7g(x)^6 g'(x). \] Now we need to find \( g'(x) \). We can use the quotient rule, which states that if \( g(x) = \frac{a(x)}{b(x)} \), then \[ g'(x) = \frac{a'(x)b(x) - a(x)b'(x)}{(b(x))^2}. \] Here, \( a(x) = x + 4 \) and \( b(x) = x + 6 \). Thus, \[ a'(x) = 1 \quad \text{and} \quad b'(x) = 1. \] Applying the quotient rule: \[ g'(x) = \frac{(1)(x+6) - (x+4)(1)}{(x+6)^2} = \frac{x + 6 - x - 4}{(x+6)^2} = \frac{2}{(x+6)^2}. \] Now substituting \( g'(x) \) back into \( v'(x) \): \[ v'(x) = 7 \left( \frac{x+4}{x+6} \right)^6 \cdot \frac{2}{(x+6)^2} = \frac{14 \left( x+4 \right)^6}{(x+6)^8}. \] Now we can apply the product rule: \[ f'(x) = u'v + uv' = (3x^2) \left( \frac{x+4}{x+6} \right)^7 + x^3 \left( \frac{14 \left( x+4 \right)^6}{(x+6)^8} \right). \] Now we can combine these two terms: \[ f'(x) = 3x^2 \left( \frac{x+4}{x+6} \right)^7 + \frac{14x^3 \left( x+4 \right)^6}{(x+6)^8}. \] To combine these into a single fraction, we can express the first term with a common denominator: \[ f'(x) = \frac{3x^2 (x+6)^8 (x+4)^7 + 14x^3 (x+4)^6}{(x+6)^8}. \] Thus, the derivative \( f'(x) \) is: \[ \boxed{f'(x) = \frac{3x^2 (x+4)^7 (x+6) + 14x^3 (x+4)^6}{(x+6)^8}}. \]