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\begin{lstlisting}[language=python, caption=Regresi Linier HP dan Kamera]

import numpy as np

import pandas as pd

from matplotlib import pyplot as plt

from sklearn.linear_model import LinearRegression

df = pd.DataFrame([[8,7],[2,3],[6,7],[4,2],[7,8],[3,3]])

df.columns = ['x', 'y']

x_train = df['x'].values[:,np.newaxis]

y_train = df['y'].values

lm = LinearRegression()

lm.fit(x_train,y_train) # fase training

print('Coefficient : ' + str(lm.coef_))

print('Intercept : ' + str(lm.intercept_))

x_test = [[170],[171]] # data yang akan diprediksi

p = lm.predict(x_test) # fase prediksi

print(p) # hasil prediksi

# prepare plot

pb = lm.predict(x_train)

dfc = pd.DataFrame({'x': df['x'],'y':pb})

plt.scatter(df['x'],df['y'])

plt.plot(dfc['x'],dfc['y'],color='red',linewidth=1)

plt.xlabel('kamera')

plt.ylabel('harga')

plt.show()

\end{lstlisting}

Problem Set 2.5 - Chain Rule

 16. $ y= cos^3(\frac{x^2}{1 - x}) $

$ v= \frac{x^2}{1-x}, ~ u=cos \, v, ~ y = u^3 $

$\frac{dy}{dx} = \frac{dv}{dx}. \frac{du}{dv}. \frac{dy}{du}$

$ = (\frac{(1 - x)D_x(x^2) - (x^2) D_x(1 - x)}{(1 - x)^2}). (- sin \, v).(3u^2) $

$ = \frac{(1-x)(2x) -(x^2)(-1)}{(1 - x)^2}. sin(\frac{x^2}{1 -x}). (-3 cos^2 (\frac{x^2}{1 - x}))$

$ = \frac{-3(2x - x^2)}{(1 - x)^2}cos^2(\frac{x^2}{1-x}) sin(\frac{x^2}{1 -x }) $

Problem Set 2.4 Derivatif Trigonometri

catatan :

$ \frac{d}{dx} sin x~= cos\, x$

$ D_x (cos x)~= - sin\, x$

$ \frac{d}{dx}tanx~=sec^2 x$

$D_x (cot\, x)~=-csc\, x $

$\frac{d}{dx} sec x~=sec\, x\, tan\, x$ 

--------

$ 6. ~y = csc\, x$

$ D_x(csc \, x) = D_x \frac{1}{sin x}$

$ = \frac{sin x\, D_x(1) - (1)\, D_x(sin x)}{sin^2 x}$

$ = \frac{0 - cos\, x}{sin^2 x}~=\frac{-1}{sin x}.\frac{cos\, x}{sin\, x}~= -csc\,x \,cot\, x$

$10.~y= \frac{sin x + cos x}{tan x} $ 

$\frac{d}{dx}(\frac{sin x + cos x}{tan x})$

$=\frac{tanx\, D_x (sin x+ cos x) - (sin x+ cos x)\, D_x (tan x)}{tan^2 x}$

 $= \frac{tanx(cosx-sinx)-(sinx+cosx)sec^2 x}{tan^2x}$

$= \frac{tanx(cosx-sinx)-(sinx+cosx)sec^2 x}{\frac{sin^2x}{cos^2x}}$ 

$ = [sin x - \frac{sin^2 x}{cos x} - \frac{sin x}{cos^2 x} - \frac{1}{cos x}]  : [\frac{sin^2 x}{cos^2 x}]$

$ = [sin x - \frac{sin^2 x}{cos x} - \frac{sin x}{cos^2 x} - \frac{1}{cos x}]  [\frac{cos^2 x}{sin^2 x}]$

$ =\frac{cos^2 x}{sin x} - cos x - \frac{1}{sin x} - \frac{cos x}{sin^2 x} $