The function given in the image is:

$f(x) = e^x \left( x^2 + 4 \right)(x + 6)$

We are tasked with calculating the derivative $f'(x)$.

Step-by-step solution:

1. Expand the terms inside the parentheses: First, expand $(x^2 + 4)(x + 6)$.

$(x^2 + 4)(x + 6) = x^2(x + 6) + 4(x + 6) = x^3 + 6x^2 + 4x + 24$

Now, the function becomes:

$f(x) = e^x (x^3 + 6x^2 + 4x + 24)$

2. Apply the product rule: The product rule states that if $f(x) = u(x) v(x)$, then $f'(x) = u'(x) v(x) + u(x) v'(x)$. Here, $u(x) = e^x$ and $v(x) = x^3 + 6x^2 + 4x + 24$.

   • The derivative of $u(x) = e^x$ is $u'(x) = e^x$.

   • Now, calculate the derivative of $v(x) = x^3 + 6x^2 + 4x + 24$:

     $v'(x) = 3x^2 + 12x + 4$

3. Use the product rule formula:

$f'(x) = u'(x) v(x) + u(x) v'(x)$ Substitute the known values:

$f'(x) = e^x (x^3 + 6x^2 + 4x + 24) + e^x (3x^2 + 12x + 4)$

4. Factor out $e^x$:

$f'(x) = e^x \left( (x^3 + 6x^2 + 4x + 24) + (3x^2 + 12x + 4) \right)$

5. Simplify the terms inside the parentheses:

$f'(x) = e^x \left( x^3 + 9x^2 + 16x + 28 \right)$

Thus, the derivative is:

$f'(x) = e^x (x^3 + 9x^2 + 16x + 28)$

Final Answer:

$f'(x) = e^x (x^3 + 9x^2 + 16x + 28)$

Would you like further details on any of the steps or an alternative approach?

Here are 5 related questions:

1. How do you find the derivative of a product of three or more functions?
2. What is the significance of the exponential function $e^x$ in calculus?
3. How does the chain rule differ from the product rule in derivative calculations?
4. What are some common mistakes to avoid when applying the product rule?
5. How can implicit differentiation be used when functions are not explicitly solved for one variable?

Tip: When dealing with a product involving an exponential function, always check if you can simplify terms first before applying the product rule. This makes the differentiation process smoother!