Differentiate f(x) = 𝑥3 +1 4𝑥2+1 using the quotient rule (

To differentiate the function \( f(x) = \frac{x^3 + 1}{4x^2 + 1} \) using the quotient rule, follow these steps.

### Quotient Rule Formula:
\[
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}
\]
Where:
- \( u(x) = x^3 + 1 \)
- \( v(x) = 4x^2 + 1 \)

### Step 1: Differentiate \( u(x) \) and \( v(x) \)
- \( u'(x) = \frac{d}{dx}(x^3 + 1) = 3x^2 \)
- \( v'(x) = \frac{d}{dx}(4x^2 + 1) = 8x \)

### Step 2: Apply the Quotient Rule
\[
f'(x) = \frac{(4x^2 + 1) \cdot (3x^2) - (x^3 + 1) \cdot (8x)}{(4x^2 + 1)^2}
\]

### Step 3: Simplify the numerator
\[
f'(x) = \frac{(4x^2 + 1)(3x^2) - (x^3 + 1)(8x)}{(4x^2 + 1)^2}
\]
Expanding both terms:
- \( (4x^2 + 1)(3x^2) = 12x^4 + 3x^2 \)
- \( (x^3 + 1)(8x) = 8x^4 + 8x \)

Now subtract:
\[
f'(x) = \frac{(12x^4 + 3x^2) - (8x^4 + 8x)}{(4x^2 + 1)^2}
\]
Simplifying:
\[
f'(x) = \frac{4x^4 + 3x^2 - 8x}{(4x^2 + 1)^2}
\]

Thus, the derivative of \( f(x) = \frac{x^3 + 1}{4x^2 + 1} \) is:
\[
f'(x) = \frac{4x^4 + 3x^2 - 8x}{(4x^2 + 1)^2}
\]

Would you like a further breakdown of the steps or any clarifications?

Here are 5 related questions:
1. What is the chain rule in differentiation, and how is it applied?
2. How does the product rule differ from the quotient rule in calculus?
3. Can the quotient rule be extended to more complicated functions like trigonometric functions?
4. What are the conditions under which the quotient rule is undefined?
5. How would you integrate \( f(x) = \frac{x^3 + 1}{4x^2 + 1} \)?

**Tip:** Always check if the function can be simplified before applying the quotient rule to reduce errors.

Differentiate f(x) = x^3 + 1 / 4x^2 + 1 using the quotient rule.

Explore how to differentiate the function f(x) = (x^3 + 1) / (4x^2 + 1) using the quotient rule. This step-by-step solution covers key concepts in calculus, including the application of the quotient rule. Suitable for advanced calculus students in Grades 11-12.

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