The problem is to find the derivative of the function $y = \frac{1}{3}x(\log x - 1)$ with respect to $x$. We will assume the base of the logarithm is 10.

1. Problem Description

The problem is to find the derivative of the function $y = \frac{1}{3}x(\log x - 1)$ with respect to $x$. We will assume the base of the logarithm is

1

0.

2. Solution Steps

To find the derivative of $y = \frac{1}{3}x(\log x - 1)$, we use the product rule and the derivative of $\log x$.

The product rule states that if $y = uv$, then $\frac{dy}{dx} = u'v + uv'$.

Here, let $u = \frac{1}{3}x$ and $v = \log x - 1$.

Then $u' = \frac{d}{dx}(\frac{1}{3}x) = \frac{1}{3}$ and $v' = \frac{d}{dx}(\log x - 1) = \frac{d}{dx}(\log x) - \frac{d}{dx}(1)$.

We know that the derivative of $\log_{10} x$ is $\frac{1}{x \ln 10}$. Thus, $v' = \frac{1}{x \ln 10} - 0 = \frac{1}{x \ln 10}$.

Now, applying the product rule:

$\frac{dy}{dx} = u'v + uv' = \frac{1}{3}(\log x - 1) + \frac{1}{3}x(\frac{1}{x \ln 10}) = \frac{1}{3}\log x - \frac{1}{3} + \frac{1}{3 \ln 10}$.

We can factor out $\frac{1}{3}$:

$\frac{dy}{dx} = \frac{1}{3}(\log x - 1 + \frac{1}{\ln 10})$.

Since $1 = \log 10$, we have $\frac{dy}{dx} = \frac{1}{3}(\log x - \log 10 + \frac{1}{\ln 10})$.

Since $\frac{1}{\ln 10}$ is approximately $0.434$, we can leave the expression as:

$\frac{dy}{dx} = \frac{1}{3}\log x - \frac{1}{3} + \frac{1}{3 \ln 10} = \frac{1}{3} (\log x - 1 + \frac{1}{\ln 10})$.

3. Final Answer

$\frac{dy}{dx} = \frac{1}{3}(\log x - 1) + \frac{1}{3 \ln 10} = \frac{1}{3}(\log x - 1 + \frac{1}{\ln 10})$