Consider the following statements, where x > 0:
(A) \(\displaystyle\frac{d}{dx}\left(e^{-2x}\right) = -2e^{-2x}\)
(B) \(\displaystyle\frac{d}{dx}\left(3^{x}\right) = x \cdot 3^{x - 1}\)
(C) \(\displaystyle\frac{d}{dx}\left(\log_e x^{2}\right) = \frac{2}{x}\)
(D) \(\displaystyle\frac{d}{dx}\left(\log_{10} x\right) = \frac{1}{x}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (C) only
The standard results needed are:
\( \displaystyle \frac{d}{dx}e^{kx} = k\,e^{kx} \)
\( \displaystyle \frac{d}{dx}a^{x} = a^{x}\log_e a \)
For x > 0, \(\log_e x^{2} = 2\log_e x\). By the change of base, \(\log_{10} x = \frac{\log_e x}{\log_e 10}\):
\( \displaystyle \frac{d}{dx}\left(2\log_e x\right) = \frac{2}{x} \)
\( \displaystyle \frac{d}{dx}\left(\log_{10} x\right) = \frac{1}{x\log_e 10} \)
So the derivatives are \(-2e^{-2x}\), \(3^{x}\log_e 3\), \(\frac{2}{x}\) and \(\frac{1}{x\log_e 10}\).
Checking the statements
(A) \(\frac{d}{dx}\left(e^{-2x}\right) = -2e^{-2x}\) — Correct. Here k = −2.
(B) \(\frac{d}{dx}\left(3^{x}\right) = x \cdot 3^{x - 1}\) — Incorrect. The derivative is \(3^{x}\log_e 3\); the power rule needs the variable in the base.
(C) \(\frac{d}{dx}\left(\log_e x^{2}\right) = \frac{2}{x}\) — Correct. \(\log_e x^{2} = 2\log_e x\).
(D) \(\frac{d}{dx}\left(\log_{10} x\right) = \frac{1}{x}\) — Incorrect. The derivative is \(\frac{1}{x\log_e 10}\).