If \( y=7^{x}\log(5x^{3}) \), then \( \displaystyle\frac{dy}{dx} \) is equal to: (consider \( \log_{e}x=\log x \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 7^{x}\left[\log 7\,\{\log(5x^{3})\}+\displaystyle\frac{3}{x}\right] \)
By the product rule, with
\( \displaystyle \frac{d}{dx}\left(7^{x}\right)=7^{x}\log 7 \)
\( \displaystyle \frac{d}{dx}\log(5x^{3})=\frac{15x^{2}}{5x^{3}}=\frac{3}{x} \)
we get
\( \displaystyle \frac{dy}{dx}=7^{x}\log 7\cdot\log(5x^{3})+7^{x}\cdot\frac{3}{x} \)
\( \displaystyle =7^{x}\left[\log 7\,\{\log(5x^{3})\}+\frac{3}{x}\right] \)
Tip: \( \frac{d}{dx}a^{x}=a^{x}\log a \), not \( x\,a^{x-1} \): the power rule works only when the base is the variable. Options (1) and (2) apply the power rule to \( 7^{x} \).