9. find the slope of the tangent line to f(x) = (e^(x + 2))/x^2 at x = 2. 10. the monthly revenue of new…

9.

Explanation:

Step1: Apply quotient - rule for differentiation

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = e^{x + 2}$ and $v=x^{2}$. First, find $u^\prime$ and $v^\prime$. The derivative of $u = e^{x+2}$ with respect to $x$ is $u^\prime=e^{x + 2}$ (since the derivative of $e^{ax + b}$ is $ae^{ax + b}$ with $a = 1$ and $b = 2$), and $v^\prime = 2x$. Then $f^\prime(x)=\frac{e^{x + 2}\cdot x^{2}-e^{x + 2}\cdot2x}{x^{4}}=\frac{e^{x + 2}(x^{2}-2x)}{x^{4}}$.

Step2: Evaluate the derivative at $x = 2$

Substitute $x = 2$ into $f^\prime(x)$. $f^\prime(2)=\frac{e^{2 + 2}(2^{2}-2\times2)}{2^{4}}=\frac{e^{4}(4 - 4)}{16}=0$.

Answer:

$0$

10.

Explanation:

Step1: Differentiate the revenue function

The revenue function is $R(t)=500t-10t^{2}$. Using the power - rule for differentiation, if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. The derivative of $R(t)$ with respect to $t$ is $R^\prime(t)=500-20t$.

Step2: Evaluate the derivative at $t = 8$

Substitute $t = 8$ into $R^\prime(t)$. $R^\prime(8)=500-20\times8=500 - 160 = 340$.

Answer:

$340$ dollars per sound - bar per month

11.

Explanation:

Step1: Find the total spent in 2015

Since $t$ is the number of years since 2010, for 2015, $t = 5$. Substitute $t = 5$ into $S(t)=34\ln t+155$. $S(5)=34\ln(5)+155\approx34\times1.6094 + 155=54.7196+155 = 209.7196$ (billions of dollars).

Step2: Differentiate the function $S(t)$

The derivative of $S(t)=34\ln t+155$ with respect to $t$ is $S^\prime(t)=\frac{34}{t}$ (using the rule that the derivative of $\ln x$ is $\frac{1}{x}$).

Step3: Evaluate the derivative at $t = 5$

Substitute $t = 5$ into $S^\prime(t)$. $S^\prime(5)=\frac{34}{5}=6.8$ (billions of dollars per year).

Answer:

The total spent in 2015 was approximately $209.72$ billions of dollars and it was increasing at a rate of $6.8$ billions of dollars per year.