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help me solve this | 9 parts remaining
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
use the formula for instantaneous rate of change.
lim_{h→0} (f(x + h)-f(x))/h
Step1: First find $f(x + h)$
Given $f(x)=3x^{x}$, then $f(x + h)=3(x + h)^{x + h}$
Step2: Substitute into the limit formula
The limit formula for instantaneous - rate of change is $\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{3(x + h)^{x + h}-3x^{x}}{h}$. We are interested in $x = 3$, so we substitute $x = 3$ into the formula:
$\lim_{h
ightarrow0}\frac{3(3 + h)^{3 + h}-3\times3^{3}}{h}=\lim_{h
ightarrow0}\frac{3(3 + h)^{3 + h}-81}{h}$
We can also use the logarithmic - differentiation method. Let $y = 3x^{x}$, then $\ln y=\ln(3x^{x})=\ln3+x\ln x$.
Differentiating both sides with respect to $x$:
$\frac{y^\prime}{y}=0+\ln x + x\times\frac{1}{x}=\ln x + 1$
So $y^\prime=y(\ln x + 1)=3x^{x}(\ln x + 1)$
Step3: Evaluate at $x = 3$
Substitute $x = 3$ into $y^\prime$:
$y^\prime|_{x = 3}=3\times3^{3}(\ln3 + 1)=81(\ln3 + 1)$
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$81(1+\ln3)$