The problem asks us to find the derivative of the function $f(x) = 3x^3 (4^x)$ at $x = 1$.

Step-by-Step Solution:

1. Identify the Function and Apply Product Rule: The function $f(x) = 3x^3 \cdot 4^x$ is a product of two functions:

   • $u(x) = 3x^3$
   • $v(x) = 4^x$

   We can use the product rule for differentiation: $f'(x) = u'(x)v(x) + u(x)v'(x)$

2. Differentiate Each Part:

   • For $u(x) = 3x^3$: $u'(x) = 3 \cdot 3x^2 = 9x^2$
   • For $v(x) = 4^x$: $v'(x) = 4^x \ln(4)$ (since the derivative of $a^x$ is $a^x \ln(a)$).

3. Apply the Product Rule: Substitute $u(x), u'(x), v(x),$ and $v'(x)$ into the product rule formula: $f'(x) = (9x^2)(4^x) + (3x^3)(4^x \ln(4))$ Simplify by factoring out $4^x$: $f'(x) = 4^x (9x^2 + 3x^3 \ln(4))$

4. Evaluate at $x = 1$: Substitute $x = 1$ into $f'(x)$: $f'(1) = 4^1 (9 \cdot 1^2 + 3 \cdot 1^3 \ln(4))$ Simplify further: $f'(1) = 4 (9 + 3 \ln(4))$ $f'(1) = 36 + 12 \ln(4)$

Answer:

The correct answer is: $\boxed{\text{a. } f'(1) = 12 \ln 4 + 36}$

Would you like further details on the steps, or have any other questions?

Related Questions

1. How would you apply the product rule if the function had three terms instead of two?
2. What is the derivative of $4^x$ with respect to $x$?
3. How would the answer change if $f(x) = 3x^3 e^x$ instead?
4. Why is the natural logarithm used in the derivative of an exponential function?
5. How do you evaluate $f'(x)$ if given $x = 2$ instead?

Tip

Always look for common factors in derivatives to simplify expressions effectively, especially when applying the product rule.