On section 4 of part A of the first unit (wow that is a mouthful) how are we expected to know how to calculate the derivative of sin 2x? On the worked example, we need to know this somehow.
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You are not. To calculate the derivative of sin2x you need to know the derivative of sinx and you need to know about the chain rule. Both are explained later in the course (sessions 7 and 11 respectively). It's a little strange that this problem is included before those sessions but if you look at them you should be able to understand it.
Method 1: By Chain Rule
If y = sin 2x then let u = 2x and then y = sin u. Now
du/dx = 2 and dy/du = cos u [∵ the graph of the derivative (slope) of the sine function is the cosine function].
By chain rule in Leibniz notation,
dy/dx = (dy/du)(du/dx)
dy/dx = 2(cos u)
Replacing u with 2x, we obtain
dy/dx = 2(cos 2x)
Method 2: Taking the derivative of the six trigonometric functions.
d(sin x)/dx = cos x
d(cos x)/dx = -sin x
d(tan x)/dx = sec²x
d(csc x)/dx = -(csc x)(cot x)
d(sec x)/dx = (sec x)(tan x)
Rule: To differentiate on of the six trigonometric functions:
i). multiply the derivative of the angle by the amplitude,
ii). multiply this product by the derivative of the respective trig function. Remember to never change the angle.
Thus if y = sin 2x then
dy/dx = [d(2x)/dx](1)(cos, the derivative of sin, and never change the angle)
∴ dy/dx = 2(cos 2x)
Note you may also find the derivative from the definition of the derivative.
Watch the recitation video for the class titled "Derivatives of Sine and Cosine". Its available for download at http://archive.org/details/MIT18_01SCF10/ or you can watch it online at http://www.youtube.com/watch?v=Bb-bgJdOqig&feature=plcp