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. \( \frac{d}{d x}(\ln k x)=\frac{d}{d x} \)

To show that \( \frac{d}{d x}(\ln k x) = \frac{d}{d x} \ln x \), let's first compute the derivative of \( \ln k x \). Using the property of logarithms, we can rewrite \( \ln k x \) as: \[ \ln k x = \ln k + \ln x. \] Then, when we differentiate, the derivative of a constant (like \( \ln k \)) is zero, so we get: \[ \frac{d}{d x}(\ln k x) = \frac{d}{d x}(\ln k + \ln x) = 0 + \frac{d}{d x} \ln x = \frac{1}{x}. \] So, we have shown that \( \frac{d}{d x}(\ln k x) = \frac{1}{x} \). Now to compute the derivative of \( \ln(k x) \) while letting \( u = k x \). Using the chain rule: \[ \frac{d}{d x}(\ln u) = \frac{1}{u} \cdot \frac{d u}{d x}. \] Calculating \( \frac{d u}{d x} \): \[ u = kx \implies \frac{d u}{d x} = k. \] Thus, we have: \[ \frac{d}{d x}(\ln(k x)) = \frac{1}{k x} \cdot k = \frac{1}{x}. \] Therefore, we see that both approaches agree: \[ \frac{d}{d x}(\ln(k x)) = \frac{d}{d x}(\ln x). \] Conclusion: \( \frac{d}{d x}(\ln k x) = \frac{1}{x} \), confirming that \( \frac{d}{d x}(\ln k x) = \frac{d}{d x} \ln x \).