Question

use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller values of h, to find the instantaneous rate of change for the given function at the given value.
f(x)=3x^x; x = 3
the instantaneo
(round to three
use the formula for instantaneous rate of change. find the limit by substituting progressively smaller values of h (such as 1, .01, .001) until two evaluations yield the same answer. the instantaneous rate of change is that answer. it is also possible to use the table feature on your graphing calculator to evaluate.
\\(\lim_{h\to0}\frac{f(x + h)-f(x)}{h}\\)

Explanation:

Step1: Write the function and the point

The function is $f(x)=3x^{x}$ and $x = 3$.

Step2: Calculate $f(x + h)$

$f(3 + h)=3(3 + h)^{3 + h}=3(3 + h)^{3}(3 + h)^{h}$.

Step3: Substitute into the limit formula

The instantaneous - rate - of - change formula is $\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{3(3 + h)^{3 + h}-3\times3^{3}}{h}$.
We can also use the logarithmic differentiation method. Let $y = 3x^{x}$, then $\ln y=\ln(3)+x\ln x$.
Differentiating both sides with respect to $x$:
$\frac{y'}{y}=0+\ln x + x\times\frac{1}{x}=\ln x + 1$.
So $y'=y(\ln x + 1)=3x^{x}(\ln x + 1)$.

Step4: Evaluate at $x = 3$

Substitute $x = 3$ into $y'$:
$y'(3)=3\times3^{3}(\ln3 + 1)=81(\ln3 + 1)$.
Using $\ln3\approx1.099$, we have $y'(3)=81\times(1.099 + 1)=81\times2.099 = 169.019$.

Answer:

$169.019$