anonymous
  • anonymous
In Problem Set 1: Problem 1J-2: How does d/dx cos(x) when x is PI/2 = Cos(x)-Cos(PI/2)/(x-PI/2). Shouldn't the numerator be Cos(x-PI/2)-Cos(PI/2)? (Difference Quotient)
MIT 18.01 Single Variable Calculus (OCW)
chestercat
  • chestercat
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NoelGreco
  • NoelGreco
Some people who might want to help you may not know how to find problem 1J-2. Posting a link might get you more help.
anonymous
  • anonymous
Thanks for the advice. http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/problem-set-1/MIT18_01SC_pset1prb.pdf
anonymous
  • anonymous
We define derivative this way: \[ f'(x) = \lim_{\Delta x \rightarrow 0 } \frac { f(x + \Delta x) - f(x) } { \Delta x } \] Substitute \[ \Delta x = x_0 - x \] to get \[ f'(x) = \lim_{ x \rightarrow x_0 } \frac { f(x + (x_0 - x)) - f(x) } { x_0 - x } \\ = \lim_{ x \rightarrow x_0 } \frac { f(x_0) - f(x) } { x_0 - x } \\ = \lim_{ x \rightarrow x_0 } \frac { - (f(x) - f(x_0)) } { -(x - x_0) } \\ = \lim_{ x \rightarrow x_0 } \frac { f(x) - f(x_0) } { x - x_0 } \] Therefore \[ \cos'(x) = \lim_{x \rightarrow \pi/2} \frac { \cos ( x ) - \cos (\pi /2) } { x - \pi/2} = \lim_{x \rightarrow \pi/2} \frac { \cos ( x ) - \cos( \pi /2) } { x - \pi/2} \]

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