kikiTauNerds
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The question for Session4 requires you to find the derivative of sin(2x). I tried using various sin/cos properties but nothing was able to cancel out the deltaX in the denominator. I tried things like sin(x+y) = sin(x)cos(y)+cos(x)sin(y), and cos(x+y) = cos(x)cos(y)-sin(x)sin(y), and sin(x)^2 = 1 - cos(x)^2.

Answer :

HomertheGenius
By the chain rule:
( sin ( 2 x ) ) ` = cos ( 2 x ) · ( 2 x )` = 2 cos ( 2 x )
f ` ( sin 2 x )  = [tex] \lim_{h \to 0} \frac{sin(2x+2h)- sin(2x)}{h}= \\ \lim_{h \to 0} \frac{sin(2x)*sin(2h)+sin(2h)cos(2x)-sin(2x)}{h} = \\ \lim_{h \to 0} \frac{sin(2h)*cos(2x)}{h} = \\ \lim_{h \to 0} \frac{2sin(2h)*cos(2x)}{2h}= [/tex]
= 2 cos ( 2 x )
because: [tex] \lim_{x \to 0} \frac{sinx}{x} =1 [/tex]
This is the reason why Δ x ( or h here ) is cancelled in the denominator

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