For $y=x^3\, log_ex, x> 0$
A. $\left.\frac{dy}{dx}\right]_{x=e}=4e$
B. $\frac{d^2y}{dx^2}=x^2(5+log_ex^3)$
C. $\frac{dy}{dx}=x^2(1+log_ex^3)$
D. $\left.\frac{d^2y}{dx^2}\right]_{x=e}=11e$
E. $\left.\frac{d^2y}{dx^2}\right]_{x=1}=5$
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C, D, E
$y=x^3\log_ex$
$\frac{dy}{dx}=3x^2\log_ex+x^2$
$=x^2(3\log_ex+1)=x^2(\log_ex^3+1)$ → (C)
Now,
$\frac{d^2y}{dx^2}=3(2x\log_ex+x)+2x$
$\left.\frac{d^2y}{dx^2}\right|_{x=e}=3(2e\log_ee+e)+2e$
$=3(2e+e)+2e$
$=11e$ → (D)
$\left.\frac{d^2y}{dx^2}\right|_{x=1}=3(2×0+1)+2(1)$
$=5$ → (E)