Consider the following statements:
(A) \(\displaystyle\frac{d}{dx}\left(\log_{10} x\right) = \frac{1}{x\log_e 10}\), \(x \gt 0\)
(B) \(\displaystyle\frac{d}{dx}\left(\log_e|x|\right) = -\frac{1}{x}\), \(x \lt 0\)
(C) \(\displaystyle\frac{d}{dx}\left(e^{-3x}\right) = -3e^{-3x}\)
(D) \(\displaystyle\frac{d}{dx}\left(5^{x}\right) = x \cdot 5^{x - 1}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
By the change of base rule, \(\log_{10} x = \frac{\log_e x}{\log_e 10}\).
For x < 0, |x| = −x, and the chain rule gives
\( \displaystyle \frac{d}{dx}\log_e(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x} \)
For an exponential function with a constant base, \(\frac{d}{dx}a^{x} = a^{x}\log_e a\).
So the derivatives are \(\frac{1}{x\log_e 10}\), \(\frac{1}{x}\), \(-3e^{-3x}\) and \(5^{x}\log_e 5\).
Checking the statements
(A) \(\frac{d}{dx}\left(\log_{10} x\right) = \frac{1}{x\log_e 10}\), \(x \gt 0\) — Correct. It is \(\frac{1}{\log_e 10}\) times the derivative \(\frac{1}{x}\) of \(\log_e x\).
(B) \(\frac{d}{dx}\left(\log_e|x|\right) = -\frac{1}{x}\), \(x \lt 0\) — Incorrect. The derivative of \(\log_e(-x)\) is \(\frac{1}{x}\), with no minus sign.
(C) \(\frac{d}{dx}\left(e^{-3x}\right) = -3e^{-3x}\) — Correct. Chain rule: \(e^{-3x} \times (-3)\).
(D) \(\frac{d}{dx}\left(5^{x}\right) = x \cdot 5^{x - 1}\) — Incorrect. The power rule does not apply when x is in the power; the derivative is \(5^{x}\log_e 5\).