Math Problem Statement

find f''(x) f(x)=ln[(x2+4)(x3+8x−3)2]

Solution

We are given the function f(x)=ln[(x2+4)(x3+8x3)2]f(x) = \ln\left[(x^2 + 4)(x^3 + 8x - 3)^2\right]. To find the second derivative f(x)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)\ln(ab) = \ln(a) + \ln(b) and ln(b2)=2ln(b)\ln(b^2) = 2\ln(b)

Thus, the given function becomes:

f(x)=ln(x2+4)+2ln(x3+8x3)f(x) = \ln(x^2 + 4) + 2\ln(x^3 + 8x - 3)

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

Now, differentiate f(x)f(x) with respect to xx 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)=1x2+4ddx(x2+4)+21x3+8x3ddx(x3+8x3)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:

ddx(x2+4)=2x\frac{d}{dx}(x^2 + 4) = 2x and ddx(x3+8x3)=3x2+8\frac{d}{dx}(x^3 + 8x - 3) = 3x^2 + 8

Thus, we have:

f(x)=2xx2+4+23x2+8x3+8x3f'(x) = \frac{2x}{x^2 + 4} + 2 \cdot \frac{3x^2 + 8}{x^3 + 8x - 3}

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

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

ddx(uv)=vuuvv2\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:

ddx(2xx2+4)\frac{d}{dx} \left( \frac{2x}{x^2 + 4} \right) Here, u=2xu = 2x and v=x2+4v = x^2 + 4. Differentiating:

u=2u' = 2 and v=2xv' = 2x

Applying the quotient rule:

= \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.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Logarithmic Functions
Chain Rule
Product Rule
Quotient Rule

Formulas

ln(ab) = ln(a) + ln(b)
ln(b^2) = 2ln(b)
Quotient Rule: d/dx(u/v) = (v * u' - u * v') / v^2
Chain Rule: d/dx[f(g(x))] = f'(g(x)) * g'(x)

Theorems

Logarithmic Differentiation Theorem
Quotient Rule Theorem
Chain Rule Theorem

Suitable Grade Level

Undergraduate Calculus