To find the second derivative of $f(x) = \frac{1}{(x^2 + 4)^2}$, let's go step by step.

Step 1: First Derivative

We can use the chain rule and quotient rule to find the first derivative.

We rewrite $f(x) = (x^2 + 4)^{-2}$.

Now, differentiate using the chain rule:

$f'(x) = -2(x^2 + 4)^{-3} \cdot (2x)$

Simplifying:

$f'(x) = \frac{-4x}{(x^2 + 4)^3}$

Step 2: Second Derivative

Now, to find the second derivative $f''(x)$, we differentiate $f'(x)$.

Differentiate $f'(x) = \frac{-4x}{(x^2 + 4)^3}$ using the quotient rule.

Let:

• $u = -4x$
• $v = (x^2 + 4)^3$

Using the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$, we need to find $u'$ and $v'$.

• $u' = -4$
• $v' = 3(x^2 + 4)^2 \cdot (2x) = 6x(x^2 + 4)^2$

Now, applying the quotient rule:

$f''(x) = \frac{(-4)(x^2 + 4)^3 - (-4x)(6x(x^2 + 4)^2)}{(x^2 + 4)^6}$

Simplifying:

$f''(x) = \frac{-4(x^2 + 4)^3 + 24x^2(x^2 + 4)^2}{(x^2 + 4)^6}$

Factor out $(x^2 + 4)^2$ from the numerator:

$f''(x) = \frac{(x^2 + 4)^2 \left[ -4(x^2 + 4) + 24x^2 \right]}{(x^2 + 4)^6}$

Simplify the expression inside the brackets:

$-4(x^2 + 4) + 24x^2 = -4x^2 - 16 + 24x^2 = 20x^2 - 16$

Now the second derivative becomes:

$f''(x) = \frac{(x^2 + 4)^2 (20x^2 - 16)}{(x^2 + 4)^6}$

Canceling out $(x^2 + 4)^2$:

$f''(x) = \frac{20x^2 - 16}{(x^2 + 4)^4}$

Thus, the second derivative is:

$f''(x) = \frac{20x^2 - 16}{(x^2 + 4)^4}$

Would you like more details on any step or clarification?

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Here are 5 related questions:

1. What is the third derivative of $f(x) = \frac{1}{(x^2 + 4)^2}$?
2. How would you compute higher-order derivatives using general rules?
3. Can you find the critical points of $f(x) = \frac{1}{(x^2 + 4)^2}$?
4. How do second derivatives help identify concavity in functions?
5. What does the second derivative test indicate about local extrema?

Tip: The quotient rule can sometimes be avoided by simplifying the function into a product form, using the chain rule directly.