The second order derivative of which of the following functions is $5^x$?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{5^x}{\left(\log _e 5\right)^2}$
Option (1): $f(x)=5^x\log_e5$
$f'(x)=\frac{d}{dx}(5^x\log_e5)=(\log_e5).(5^x\log_e5)=5^x(\log_e5)^2$
$f''(x)=\frac{d}{dx}(5^x(\log_e5)^2)=(\log_e5)^2.(5^x\log_e5)=5^x(\log_e5)^3$ This is not $5^x$.
Option (2): $5^x(\log_e5)^2$
$f'(x)=\frac{d}{dx}(5^x(\log_e5)^2)=(\log_e5)^2.(5^x\log_e5)=5^x(\log_e5)^3$
$f''(x)=\frac{d}{dx}(5^x(\log_e5)^3)=(\log_e5)^3.(5^x\log_e5)=5^x(\log_e5)^4$ This is not $5^x$.
Option (3): $f(x)=\frac{5^x}{\log_e5}$
$f'(x)=\frac{d}{dx}\left(\frac{5^x}{\log_e5}\right)=\frac{1}{\log_e5}.(5^x(\log_e5)=5^x$
$f''(x)=\frac{d}{dx}(5^x)=5^x(\log_e5)$ This is not $5^x$.
Option (4):
$f(x)=\frac{5^x}{(\log_e5)^2}$
$f'(x)=\frac{d}{dx}\left(\frac{5^x}{(\log_e5)^2}\right)=\frac{1}{(\log_e5)^2}.(5^x\log_e5)=\frac{5^x}{\log_e5}$
$f''(x)=\frac{d}{dx}\left(\frac{5^x}{\log_e5}\right)=\frac{1}{\log_e5}.(5^x\log_e5)=5^x$ This is $5^x$.