Question

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 1

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