$$ - f ^ { \prime \prime } ( x ) = x [ \frac { - x } { \sqrt { ( x ^ { 2 } - 5 ) ^ { 3 } } } ] + \frac { 1 } { \sqrt { x ^ { 2 } - 5 } } ( 1 ) $$

Question

UpStudy AI · Accepted Answer

We start with the equation

$$
- f''(x) = x\left[\frac{-x}{\sqrt{(x^2-5)^3}}\right]+\frac{1}{\sqrt{x^2-5}}.
$$

**Step 1. Simplify the first term**

Multiply $$x$$ by the fraction:

$$
x\left[\frac{-x}{\sqrt{(x^2-5)^3}}\right] = -\frac{x^2}{\sqrt{(x^2-5)^3}}.
$$

Now the equation becomes

$$
- f''(x) = -\frac{x^2}{\sqrt{(x^2-5)^3}}+\frac{1}{\sqrt{x^2-5}}.
$$

**Step 2. Multiply through by $$-1$$ to solve for $$ f''(x) $$**

Multiplying both sides by $$-1$$ gives

$$
f''(x) = \frac{x^2}{\sqrt{(x^2-5)^3}}-\frac{1}{\sqrt{x^2-5}}.
$$

**Step 3. Write each term with a common denominator**

Recall that

$$
\sqrt{(x^2-5)^3} = (x^2-5)^{3/2} \quad \text{and} \quad \sqrt{x^2-5} = (x^2-5)^{1/2}.
$$

Thus, the equation is

$$
f''(x) = \frac{x^2}{(x^2-5)^{3/2}} - \frac{1}{(x^2-5)^{1/2}}.
$$

Express the second term with the denominator $$(x^2-5)^{3/2}$$ by multiplying numerator and denominator by $$(x^2-5)$$:

$$
\frac{1}{(x^2-5)^{1/2}} = \frac{x^2-5}{(x^2-5)^{3/2}}.
$$

So,

$$
f''(x) = \frac{x^2}{(x^2-5)^{3/2}} - \frac{x^2-5}{(x^2-5)^{3/2}}.
$$

**Step 4. Combine the terms**

Combine the fractions:

$$
f''(x) = \frac{x^2 - (x^2-5)}{(x^2-5)^{3/2}}.
$$

Simplify the numerator:

$$
x^2 - (x^2-5) = x^2 - x^2 + 5 = 5.
$$

Thus,

$$
f''(x) = \frac{5}{(x^2-5)^{3/2}}.
$$
$$ f''(x) = \frac{5}{(x^2 - 5)^{3/2}} $$

\( - f ^ { \prime \prime } ( x ) = x [ \frac { - x } { \sqrt { ( x ^ { 2 } - 5 ) ^ { 3 } } } ] + \frac { 1 } { \sqrt { x ^ { 2 } - 5 } } ( 1 ) \)