If $f(x) = x^3\log_e x$, Then $f''(e^2)$ is equal to

Given $f(x) = x^3\log_e x = x^3 \ln x$

Using the product rule with $u = x^3$ and $v = \ln x$:

$u' = 3x^2$

$v' = \frac{1}{x}$

$f'(x) = 3x^2 \cdot \ln x + x^3 \cdot \frac{1}{x}$

$f'(x) = 3x^2 \ln x + x^2$

$f'(x) = x^2(3\ln x + 1)$

---

Using the product rule with $u = x^2$ and $v = (3\ln x + 1)$:

$u' = 2x$

$v' = \frac{3}{x}$

$f''(x) = 2x \cdot (3\ln x + 1) + x^2 \cdot \frac{3}{x}$

$f''(x) = 2x(3\ln x + 1) + 3x$

$f''(x) = 6x\ln x + 2x + 3x$

$f''(x) = 6x\ln x + 5x$

$f''(x) = x(6\ln x + 5)$

---

At $x = e^2$:

$f''(e^2) = e^2(6\ln(e^2) + 5)$

Since $\ln(e^2) = 2\ln(e) = 2$:

$f''(e^2) = e^2(6 \times 2 + 5)$

$f''(e^2) = e^2(12 + 5)$

$f''(e^2) = 17e^2$

Therefore, $f''(e^2) = 17e^2$