# Question 17db1

Feb 22, 2018

 \qquad \qquad \qquad \qquad \qquad \qquad f'(x) = {5π}/3 cos ({πx}/6 - {2π}/3).

#### Explanation:

$\text{Let's do this step by step:}$

 \qquad \qquad f(x) = 10 sin ({πx}/6 - {2π}/3) +50 \qquad \quad => 

 f'(x) = [ 10 sin ({πx}/6 - {2π}/3) ]' + [ 50 ]' \qquad \qquad \qquad \qquad \quad "by Sum Rule" 

 = 10 [ sin ({πx}/6 - {2π}/3) ]' + 0 \qquad \qquad \qquad \qquad "by Rules for Constants" 

 = 10 cos ({πx}/6 - {2π}/3) cdot ({πx}/6 - {2π}/3)' \qquad "Chain Rule with sin" 

 = 10 cos ({πx}/6 - {2π}/3) cdot ({π}/6 x- {2π}/3)' \qquad \qquad \qquad \quad \ "rewrite a little" 

 = 10 cos ({πx}/6 - {2π}/3) cdot [ ({π}/6) - 0].\qquad \qquad \ "Rules for Constants" 

$\text{Note that there are no more differentiations asked for now in}$
$\text{the last expression.}$
$\text{This means we are at the beginning of the Simplification }$
$\text{Phase. No more derivatives now.}$

 = 10({π}/6) cos ({πx}/6 - {2π}/3) 

 = {5π}/3 cos ({πx}/6 - {2π}/3). 

$\text{This is our answer.}$

$\text{Summarizing:}$

 \qquad \qquad f(x) = 10 sin ({πx}/6 - {2π}/3) +50 \qquad \quad => 

 \qquad \qquad \qquad \qquad \qquad \qquad f'(x) = {5π}/3 cos ({πx}/6 - {2π}/3).#