Xtasy

The useful trick here is to take logs: $\large f(x) = {x}^{{x}^{2}-3x+2} \implies \ln\left(f(x)\right)=\left(x^2-3x+2\right)\ln(x)$ Taking the derivative we get: $\large \frac{f'(x)}{f(x)}= (2x-3)\ln(x)+\frac{x^2-3x+2}{x} \Rightarrow f'(x)= f(x)\left((2x-3)\ln(x)+\frac{x^2-3x+2}{x}\right)$

Using the fact that $f(3)=9$ we can deduce $f'(3)=9\left(3\ln(3)+\frac{2}{3}\right)=27\ln(3)+6 \approx \boxed{36}$

$\Large f (x)$ = $\Large e^{(x^2 - 3 x + 2) \ln x}$

$\Large f '(x)$ = [(2 x - 3) $\ln x$ + ( $x^2 - 3 x + 2)$ $\frac{1}{x}$ ] $\Large e^{(x^2 - 3 x + 2) \ln x}$

$\Large f '(3)$ = [3 $\ln 3$ + $\frac{2}{3}$ ] $\Large e^{2 \ln 3}$ = 27 $\ln 3$ + 6 = 35.66253179403896166767162139690+

Answer: $\boxed{36}$

$\frac{\partial x^{x^2-3 x+2}}{\partial x} \Rightarrow x^{x^2-3 x+2} \left(\frac{x^2-3 x+2}{x}+(2 x-3) \log (x)\right)$ .

Evaluated at $x\to 3$ gives $9 \left(\frac{2}{3}+3 \log (3)\right) \approx 35.662531794039$ .