To find the derivative of $6x^2 \cdot e^{x \cdot f(2x)}$, we will apply the product rule and the chain rule.

Let’s break it down step by step:

Step 1: Apply the Product Rule

The product rule states: $\frac{d}{dx} (u \cdot v) = u'v + uv'$ Here, let:

• $u = 6x^2$
• $v = e^{x \cdot f(2x)}$

We need to find $u'$ and $v'$.

Step 2: Find $u'$

The derivative of $u = 6x^2$ is straightforward: $u' = \frac{d}{dx} 6x^2 = 12x$

Step 3: Find $v'$

Now we need to differentiate $v = e^{x \cdot f(2x)}$. This involves applying the chain rule: $v' = \frac{d}{dx} e^{x \cdot f(2x)} = e^{x \cdot f(2x)} \cdot \frac{d}{dx} (x \cdot f(2x))$

To differentiate $x \cdot f(2x)$, use the product rule again: $\frac{d}{dx} (x \cdot f(2x)) = \frac{d}{dx} x \cdot f(2x) + x \cdot \frac{d}{dx} f(2x)$

• The derivative of $x$ is 1.
• The derivative of $f(2x)$ using the chain rule is $f'(2x) \cdot 2 = g(2x) \cdot 2$ (since $g(x) = f'(x)$).

So, we have: $\frac{d}{dx} (x \cdot f(2x)) = f(2x) + 2x \cdot g(2x)$

Thus: $v' = e^{x \cdot f(2x)} \cdot (f(2x) + 2x \cdot g(2x))$

Step 4: Combine using the Product Rule

Now, using the product rule: $\frac{d}{dx} \left( 6x^2 \cdot e^{x \cdot f(2x)} \right) = u'v + uv'$

Substituting the values we found for $u'$, $v$, and $v'$: $= 12x \cdot e^{x \cdot f(2x)} + 6x^2 \cdot e^{x \cdot f(2x)} \cdot (f(2x) + 2x \cdot g(2x))$

Final Answer:

$\frac{d}{dx} \left( 6x^2 \cdot e^{x \cdot f(2x)} \right) = 12x \cdot e^{x \cdot f(2x)} + 6x^2 \cdot e^{x \cdot f(2x)} \cdot (f(2x) + 2x \cdot g(2x))$

Let me know if you'd like to see more details or have any questions!

Related Questions:

1. How is the product rule used in differentiation?
2. Can you explain how the chain rule applies to exponential functions?
3. What is the significance of $f'(x) = g(x)$ in this problem?
4. How do you differentiate a composite function involving another function?
5. How would the derivative change if $f(x)$ was a specific function, like $\sin(x)$?

Tip:

Always check for opportunities to apply both the product and chain rules in combination, as in this problem!