use the formula for instantaneous rate of change, approximating the limit by using smaller and smaller…

Explanation:

Step1: Recall the formula for instantaneous rate of change

The instantaneous rate of change of a function $y = f(x)$ at $x=a$ is given by $\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $f(x)=7x^{x}$, $a = 2$. So we need to find $\lim_{h\rightarrow0}\frac{7(2 + h)^{2 + h}-7\times2^{2}}{h}$.

Step2: First, simplify $f(2)=7\times2^{2}=7\times4 = 28$.

Let's approximate the limit by choosing small values of $h$. When $h = 0.001$: [ \begin{align*} \frac{7(2+0.001)^{2 + 0.001}-28}{0.001}&=\frac{7\times2.001^{2.001}-28}{0.001}\ 2.001^{2.001}&=2.001^{2}\times2.001^{0.001}\ 2.001^{2}&=4.004001\ 2.001^{0.001}&\approx1 + 0.001\ln(2.001)\approx1+0.001\times\ln(2)\approx1 + 0.001\times0.693147 = 1.000693147\ 2.001^{2.001}&\approx4.004001\times1.000693147\approx4.006796\ 7\times2.001^{2.001}&\approx7\times4.006796 = 28.047572\ \frac{7\times2.001^{2.001}-28}{0.001}&=\frac{28.047572 - 28}{0.001}=47.572 \end{align*} ] When $h=- 0.001$: [ \begin{align*} \frac{7(2 - 0.001)^{2-0.001}-28}{-0.001}&=\frac{7\times1.999^{1.999}-28}{-0.001}\ 1.999^{1.999}&=1.999^{2}\times1.999^{- 0.001}\ 1.999^{2}&=3.996001\ 1.999^{-0.001}&\approx1-0.001\ln(1.999)\approx1-0.001\times\ln(2)\approx1-0.001\times0.693147=0.999307\ 1.999^{1.999}&\approx3.996001\times0.999307\approx3.994803\ 7\times1.999^{1.999}&\approx7\times3.994803=27.963621\ \frac{7\times1.999^{1.999}-28}{-0.001}&=\frac{27.963621 - 28}{-0.001}=36.379 \end{align*} ] Averaging these two - sided approximations (a better approximation as $h\rightarrow0$), we get $\frac{47.572+36.379}{2}=41.976$.

Answer:

$41.976$