Q.(MCQ) The derivative of $\ln(5x)$ with respect to $x$ is:

Using the chain rule on $\ln(g(x))$ with $g(x)=5x$ gives $\dfrac{g'(x)}{g(x)}=\dfrac{5}{5x}$, which simplifies to $\dfrac{1}{x}$.

Given: find $\dfrac{d}{dx}\ln(5x)$ and choose the correct option among (a)-(d).

Step 1 — Apply the chain rule for logarithms: $\dfrac{d}{dx}\ln(g(x)) = \dfrac{g'(x)}{g(x)}$, with $g(x)=5x$, so $g'(x)=5$.

Step 2 — Substitute: $\dfrac{d}{dx}\ln(5x) = \dfrac{5}{5x} = \dfrac{1}{x}$, which is option (a).

Step 3 — Why the other options are wrong: (b) $\dfrac{5}{x}$ mistakenly skips cancelling the $5$ in the numerator with the $5$ inside $5x$ in the denominator — it treats the chain-rule numerator and the $5x$ denominator as though they don't share a common factor. (c) $\dfrac{\ln5}{x}$ confuses $\ln(5x)$ with $\log_a x$-style constant-base differentiation, which does not apply here since the base of $\ln$ is $e$, not $5$. (d) $5$ mistakenly differentiates only the inner function $5x$ (getting $g'(x)=5$) and forgets to divide by $g(x)$ at all, ignoring the logarithm's own derivative rule entirely.

Check (independent method — logarithm property): $\ln(5x) = \ln5+\ln x$ (product-to-sum property of logs), and $\ln5$ is a constant. Differentiating this rewritten form: $\dfrac{d}{dx}(\ln5+\ln x) = 0+\dfrac1x = \dfrac1x$ — identical to the chain-rule result, independently confirming option (a) without needing the chain rule at all.

Selecting (b) $\dfrac{5}{x}$ by forgetting that the $5$ from differentiating the inner function $5x$ cancels against the $5$ already inside $5x$ in the denominator of the chain-rule fraction.