MIT18 01SCF10 Ses7a - Mit ocw PDF

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Derivative of sin x, Algebraic Proof A specific derivative formula tells us how to take the derivative of a specific function: if f (x) = x n then f � (x) = nx n−1. We’ll now compute a specific formula for the derivative of the function sin x. As before, we begin with the definition of the derivative: d sin(x + ∆x) − sin(x) sin x = lim ∆x→0 dx ∆x You may remember the following angle sum formula from high school: sin(a + b) = sin(a) cos(b) + sin(b) cos(a) This lets us untangle the x from the ∆x as follows: d sin x cos ∆x + cos x sin ∆x − sin(x) sin x = lim . ∆x→0 ∆x dx We can simplify this expression using some basic algebraic facts: d sin x dx

= = =

d sin x dx

=

� � sin x cos ∆x − sin x cos x sin ∆x + ∆x→0 ∆x ∆x �sin x(cos ∆x − 1) cos x sin ∆x � + lim ∆x→0 ∆x ∆x � �cos ∆x − 1 � �sin ∆x �� lim sin x + cos x ∆x→0 ∆x ∆x � cos ∆x − 1 � � sin ∆x � lim sin x + lim cos x ∆x→0 ∆x→0 ∆x ∆x lim

We now have two familiar functions – sin x and cos x – and two ugly looking fractions to deal with. The fractions may be familiar from our discussion of removable discontinuities. lim

cos ∆x − 1 ∆x sin ∆x lim ∆x→0 ∆x

∆x→0

=

0

=

1.

Using these (as yet unproven) facts, lim sin x

∆x→0

� cos ∆x − 1 � ∆x

+ lim cos x ∆x→0

simplifies to sin x · 0 + cos x · 1 = cos x. We conclude: d sin x = cos x dx

1

� sin ∆x � ∆x

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18.01SC Single Variable Calculus�� Fall 2010 ��

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