Problem 12 - Calculus Early Transcendentals

Question

Evaluate the following derivatives. $$\frac{d}{d x}\left(\ln ^{3}\left(3 x^{2}+2\right)\right)$$

Step-by-Step Solution

Answer

Question: Find the derivative of the function \(f(x) = \ln^3(3x^2 + 2)\). Answer: The derivative of the function is \(f'(x) = \frac{18x \cdot \ln^2(3x^2 + 2)}{3x^2+2}\).

Step 1: Identify the Outer and Inner Functions

The given function can be represented as a composition of two functions: $$f(x) = u(v(x))$$ where - \(u(v) = \ln^3(v)\) is the outer function, and - \(v(x) = 3x^2 + 2\) is the inner function. Now we'll calculate the derivatives of these functions.

Step 2: Find the Derivatives of the Outer and Inner Functions

To find \(u'(v)\), we'll first take the derivative of \(u(v) = \ln^3(v)\) with respect to \(v\): $$u'(v) = \frac{d}{dv}(\ln^3(v)) = 3\ln^2(v) \cdot \frac{1}{v}$$ To find \(v'(x)\), we'll take the derivative of \(v(x) = 3x^2 + 2\) with respect to \(x\): $$v'(x) = \frac{d}{dx}(3x^2 + 2) = 6x$$

Step 3: Apply the Chain Rule

Now that we have the derivatives of the outer and inner functions, we can use the chain rule to find the derivative of the composite function \(f(x) = \ln^3(3x^2 + 2)\). According to the chain rule, we have: $$f'(x) = u'(v(x)) \cdot v'(x)$$ Substitute our expressions for \(u'(v)\) and \(v'(x)\) from Step 2: $$f'(x) = (3\ln^2(3x^2 + 2) \cdot \frac{1}{3x^2 + 2}) \cdot (6x)$$

Step 4: Simplify the Result

Finally, let's simplify the expression for \(f'(x)\): $$f'(x) = 3\ln^2(3x^2 + 2) \cdot \frac{1}{3x^2 + 2} \cdot 6x$$ $$f'(x) = 18x \cdot \ln^2(3x^2 + 2) \cdot \frac{1}{3x^2 + 2}$$ So, the derivative of the given function is$$ f'(x) = \frac{18x \cdot \ln^2(3x^2 + 2)}{3x^2+2}$$.