https://file.aiquickdraw.com/mathbot/file/1727343990780ubo6kczf.jpg

The problem provided in the image gives the function:

\[
f(x) = 2x^4 - 4x^2 - 3
\]

We are asked to find the derivative \( f'(x) \).

### Step 1: Find the derivative of each term.
1. The derivative of \( 2x^4 \) is:
   \[
   \frac{d}{dx}(2x^4) = 8x^3
   \]
2. The derivative of \( -4x^2 \) is:
   \[
   \frac{d}{dx}(-4x^2) = -8x
   \]
3. The derivative of \( -3 \) is zero, as constants vanish.

### Step 2: Combine the derivatives.
So, the derivative of the function is:

\[
f'(x) = 8x^3 - 8x
\]

### Step 3: Factor the derivative.
We can factor out \( 8x \) from the expression:

\[
f'(x) = 8x(x^2 - 1)
\]

Using the identity \( x^2 - 1 = (x - 1)(x + 1) \), the derivative becomes:

\[
f'(x) = 8x(x - 1)(x + 1)
\]

### Step 4: Match the correct answer.
From the options given, the correct one is:

\[
f'(x) = 8x(x - 1)(x + 1), \quad x \in \mathbb{R}
\]

This corresponds to the third option in the image.

---

Do you have any further questions, or would you like more details on any step? Here are some related questions you might find helpful:

1. What is the derivative of a cubic function like \( x^3 - x \)?
2. How do we apply the product rule in differentiation?
3. Can we use different techniques to simplify higher-order polynomials during differentiation?
4. How does factoring help in finding the critical points of a function?
5. What role do identities like \( x^2 - 1 = (x - 1)(x + 1) \) play in simplifying expressions?

**Tip:** Factoring is a powerful tool for simplifying polynomial derivatives and solving equations efficiently.

Se consideră funcția f: R → R, f(x) = 2x^4 - 4x^2 - 3. Derivata funcției f este egală cu?

Learn how to compute the derivative of the polynomial function f(x) = 2x^4 - 4x^2 - 3. The problem covers key calculus concepts, including the power rule for differentiation and polynomial factorization. Suitable for Grades 10-12 students.

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