Show that \( \frac{d}{d x}(\ln k x)=\frac{d}{d x} \ln x \), given that \( x>0 \) and \( k>0 \) is a real number. Compute the derivative of the natural \( \log \) of \( k x \). Begin by letting \( u=k x \), and substituting \( u \) into the equation. \[ \frac{d}{d x}(\ln k x)=\frac{d}{d x} \ln u \] Compute the derivative using the chain rule. \( \frac{d}{d x} \ln u=\square \frac{d}{d x} u \)

To show that \( \frac{d}{d x}(\ln k x) = \frac{d}{d x} \ln x \), we start by substituting \( u = k x \). Thus, we have: \[ \frac{d}{d x}(\ln(k x)) = \frac{d}{d x}(\ln(u)). \] Using the chain rule, we compute the derivative: \[ \frac{d}{d x} \ln(u) = \frac{1}{u} \frac{d u}{d x}, \] where \( \frac{d u}{d x} = k \) (since \( u = k x \)). Plugging this back in gives us: \[ \frac{d}{d x}(\ln(k x)) = \frac{1}{k x}(k) = \frac{1}{x}. \] Now, let's compute \( \frac{d}{d x} \ln(x) \): \[ \frac{d}{d x} \ln(x) = \frac{1}{x}. \] Putting it all together, we find: \[ \frac{d}{d x}(\ln k x) = \frac{1}{x} = \frac{d}{d x} \ln x. \] Thus, we have shown that \( \frac{d}{d x}(\ln k x) = \frac{d}{d x} \ln x \). To fill in the blank in the chain rule step, we have: \[ \frac{d}{d x} \ln u = \frac{1}{u} \frac{d}{d x} u = \frac{1}{k x} k. \]