Questions: Let f(x) = log3(x - 4x^2). Find f'(x). Choose 1 answer: (A) (1-8x)/(x-4x^2) ln(3) (B) (1-8x)/(x-4x^2) log3(x) (C) 1/((x-4x^2) ln(3)) (D) (1-8x)/(x-4x^2)

Transcript text: Let $f(x)=\log _{3}\left(x-4 x^{2}\right)$. Find $f^{\prime}(x)$. Choose 1 answer: (A) $\frac{1-8 x}{\left(x-4 x^{2}\right) \ln (3)}$ (B) $\frac{1-8 x}{\left(x-4 x^{2}\right) \log _{3}(x)}$ (C) $\frac{1}{\left(x-4 x^{2}\right) \ln (3)}$ (D) $\frac{1-8 x}{x-4 x^{2}}$

Solution

Solution Steps

To find the derivative $ f'(x) $ of the function $ f(x) = \log_3(x - 4x^2) $, we can use the chain rule and the properties of logarithms. First, rewrite the logarithm in terms of the natural logarithm: $ \log_3(u) = \frac{\ln(u)}{\ln(3)} $, where $ u = x - 4x^2 $. Then, apply the chain rule: the derivative of $ \ln(u) $ is $ \frac{1}{u} \cdot u' $. Calculate $ u' $ and substitute back to find $ f'(x) $.

Step 1: Rewrite the Function

We start with the function given by

\[ f(x) = \log_3(x - 4x^2). \]

Using the change of base formula for logarithms, we can rewrite this as

\[ f(x) = \frac{\ln(x - 4x^2)}{\ln(3)}. \]

Step 2: Differentiate the Function

To find the derivative $ f'(x) $, we apply the chain rule. The derivative of $ \ln(u) $ is

\[ \frac{1}{u} \cdot u', \]

where $ u = x - 4x^2 $. First, we compute $ u' $:

\[ u' = \frac{d}{dx}(x - 4x^2) = 1 - 8x. \]

Now, substituting back into the derivative formula, we have:

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