$$\text{Pg. 368 #1, 7, 15, 19, 25, 29, 37, 42, 43, 53} \\
\text{Eric Nguyen} \\
\text{14 Jan 2019}$$

$$\begin{align}
1. \quad y' &= (\ln x)7^x \\
7. \quad y' &= (\ln 7)7^{x^4+2} \cdot 4x^3 \\
15. \quad y' &= \dfrac{1}{\ln 17} \cdot \dfrac{1}{x} \\
19. \quad F'(x) &= \frac{1}{\ln{10}} \cdot \frac{1}{6x - 7} \cdot 6 \\
&= \frac{6}{(6x - 7) \ln 10} \\
25. \quad y' &= 6^x \cdot \left(\frac{d}{dx} \log_7{x}\right) + \left(\frac{d}{dx} 6^x\right) \cdot \log_7{x} \quad \text{Product rule.} \\
&= 6^x \cdot \left(\frac{1}{\ln{7}} \cdot \frac{1}{x}\right) + \left(\ln{6} \cdot 6^x\right) \cdot \log_7{x} \\
&= \frac{6^x}{x \cdot \ln{7}} + \ln{6} \cdot 6^x \cdot \log_7{x} \\
29. \quad g'(x) &= \frac{(4x + 1) \cdot \frac{d}{dx} (7^x) - \left[7^x \cdot \frac{d}{dx} (4x + 1)\right]}{(4x + 1)^2} \\
&= \frac{(4x + 1) \cdot (\ln{7} \cdot 7^x) - \left(7^x \cdot 4\right)}{(4x + 1)^2} \\
\text{37a}. \quad V'(t) &= 5200 \cdot \frac{d}{dt} (0.80)^t \\
&= 5200 \cdot \ln{0.80} \cdot (0.80)^t
\end{align}$$


```python
%matplotlib inline
%config InlineBackend.figure_format='retina'
from numpy import power, linspace, log
from matplotlib.pyplot import plot
V = lambda t: 5200 * power(0.80, t)
dV = lambda t: 5200 * log(0.80) * power(0.80, t)
xp = linspace(0, 25, 100)
_ = plot(xp, V(xp))
_ = plot(xp, dV(xp))
```

$37b. \quad V'(t) \text{ shows the rate of change in the office machine's salvage value slowly approaches 0.}$  
$42. \quad 5$  
$43. \quad 6.9$  
$53. \quad y' = \dfrac{1}{\ln 3} \cdot \dfrac{1}{\log{x}} \cdot \dfrac{1}{\ln{10}} \cdot \dfrac{1}{x}$

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ipynb
Jupyter Notebook
2019-01/2019-01-14.ipynb
airicbear/calculus-homework
a765d3ba35b2b3794b9b2cce038152682eeb2cb8
MIT
2019-01/2019-01-14.ipynb
airicbear/calculus-homework
a765d3ba35b2b3794b9b2cce038152682eeb2cb8
MIT
2019-02-04T07:00:05.000Z
2019-02-09T01:17:25.000Z
2019-01/2019-01-14.ipynb
airicbear/calculus-homework
a765d3ba35b2b3794b9b2cce038152682eeb2cb8
MIT
Qwen/Qwen-72B
1. YES
2. YES