find f''(x)
*f*(*x*)=ln[(*x*2+4)(*x*3+8*x*−3)2]

We are given the function \( f(x) = \ln\left[(x^2 + 4)(x^3 + 8x - 3)^2\right] \). To find the second derivative \( f''(x) \), we will proceed step-by-step, using the chain rule and product rule as necessary.

### Step 1: Simplify the logarithmic expression
We can apply the logarithm property:

\[
\ln(ab) = \ln(a) + \ln(b)
\]
and
\[
\ln(b^2) = 2\ln(b)
\]

Thus, the given function becomes:

\[
f(x) = \ln(x^2 + 4) + 2\ln(x^3 + 8x - 3)
\]

### Step 2: Find the first derivative \( f'(x) \)

Now, differentiate \( f(x) \) with respect to \( x \) using the chain rule.

\[
f'(x) = \frac{d}{dx}\left[\ln(x^2 + 4)\right] + 2\frac{d}{dx}\left[\ln(x^3 + 8x - 3)\right]
\]

Using the chain rule for each term:

\[
f'(x) = \frac{1}{x^2 + 4} \cdot \frac{d}{dx}(x^2 + 4) + 2 \cdot \frac{1}{x^3 + 8x - 3} \cdot \frac{d}{dx}(x^3 + 8x - 3)
\]

The derivatives of the inner functions are:

\[
\frac{d}{dx}(x^2 + 4) = 2x
\]
and
\[
\frac{d}{dx}(x^3 + 8x - 3) = 3x^2 + 8
\]

Thus, we have:

\[
f'(x) = \frac{2x}{x^2 + 4} + 2 \cdot \frac{3x^2 + 8}{x^3 + 8x - 3}
\]

### Step 3: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \) to find \( f''(x) \). This will require the quotient rule for both terms. Recall that the quotient rule is:

\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}
\]

Let's apply the quotient rule to each term.

#### First term:
\[
\frac{d}{dx} \left( \frac{2x}{x^2 + 4} \right)
\]
Here, \( u = 2x \) and \( v = x^2 + 4 \). Differentiating:

\[
u' = 2
\]
and
\[
v' = 2x
\]

Applying the quotient rule:

\[
\frac{d}{dx} \left( \frac{2x}{x^2 + 4} \right) = \frac{(x^2 + 4)(2) - (2x)(2x)}{(x^2 + 4)^2}
= \frac{2(x^2 + 4) - 4x^2}{(x^2 + 4)^2}
= \frac{-2x^2 + 8}{(x^2 + 4)^2}
= \frac{2(4 - x^2)}{(x^2 + 4)^2}
\]

#### Second term:
\[
\frac{d}{dx} \left( 2 \cdot \frac{3x^2 + 8}{x^3 + 8x - 3} \right)
\]

Here, \( u = 3x^2 + 8 \) and \( v = x^3 + 8x - 3 \). Differentiating:

\[
u' = 6x
\]
and
\[
v' = 3x^2 + 8
\]

Now applying the quotient rule:

\[
\frac{d}{dx} \left( \frac{3x^2 + 8}{x^3 + 8x - 3} \right) = \frac{(x^3 + 8x - 3)(6x) - (3x^2 + 8)(3x^2 + 8)}{(x^3 + 8x - 3)^2}
\]

This can be simplified, but the result is quite complex, so we will leave it in this form for now.

Thus, the second derivative is:

\[
f''(x) = \frac{2(4 - x^2)}{(x^2 + 4)^2} + 2 \cdot \frac{(x^3 + 8x - 3)(6x) - (3x^2 + 8)(3x^2 + 8)}{(x^3 + 8x - 3)^2}
\]

Let me know if you'd like to simplify this further or go over specific steps in more detail.

---

Would you like further clarification on this result?  
Here are five related questions to explore:

1. How do you apply the product and quotient rules in calculus?
2. What are some applications of logarithmic differentiation in solving complex functions?
3. How do higher-order derivatives relate to the curvature of a graph?
4. What techniques simplify complex logarithmic derivatives?
5. How does the chain rule apply to composite functions?

**Tip:** When dealing with logarithms, breaking them into simpler components using logarithmic properties often makes differentiation easier.

find f''(x) f(x)=ln[(x^2+4)(x^3+8x−3)^2]

Learn how to find the second derivative of the logarithmic function f(x) = ln[(x^2 + 4)(x^3 + 8x - 3)^2] using the chain rule, product rule, and quotient rule. This problem breaks down each step of differentiation and applies logarithmic properties. Suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation