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Derivatives of Sine Fucntions

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Math Portfolio Assignment: Derivatives of Sine Functions Bridget Belsher Derivatives of Sine Functions, Type I Date Set: January 14 2008 Date Submitted: February 5 2008 Father Lacombe High School Mrs. Gabel I certify this portfolio Assignment is entirely my own work The above graph, displays the unaltered sin function, in its original form. The functions domain on the graph is -2? ? 2?1. The result of this is can be seen in the two oscillations with the range. Its behavior can be determined by the behavior of the first oscillation (the oscillation starting at -2?), as the same behavior will be observed in the second oscillation (that starting at 0), due to the repetitive nature of sine functions. From the point (-2?, 0), the slope can be described as increasing with a value of one. However, one can clearly see that at ( the function has no slope, as it is a point of discontinuity, that is having a slope of zero, as the line is straight. Just after that point, the slope is -1, which causes a constant decrease in the functions y-value and increase in the functions x value, which continues until the point (?, -1), at which it reaches another point of discontinuity. The derivative on sin(x) is Cos(x), which is seen in the graph below and further explained as well.


function the points at which the slope is zero are colored red. The corresponding points in Cos(x) at which the actual x-value is zero is also highlighted, but in green. In the last graph below, to the right, the two graphs are compared, at the points where the slope of sin(x) =0 and the y value of Cos(x) actually is 0. The same could be done for the maximum and minimum values at which the slope is at its greatest negative and greatest positive slope; however I think at this point the explanation is clear enough for the concept to be understood without additional graphs. When the sin function is further complicated by being multiplied by a constant, its function changes from y=sin(x) to y=aSin(x). The presence of the a-value completely changes the the appearance of the function, as well as that functions derivative, which is y=aCos(x), this change can be seen visually below and will be further explained after the graphs. The change in the function will be displayed in the example y=4Sin(x) As can be seen in the above graph, the addition of an a value cause the function to vertically stretch according to the value of a. This does not cause the y or x intercepts to change, only the absolute maximums and minimums of the function.


+ sin(2h)cos(2x) - sin(2x) h? 0 h Lim sin(2x)(cos(2h) - 1) + sin(2h)cos(2x) h? 0 h sin(2x) Lim (cos(2h) - 1) + cos(2x) Lim sin(2h) h? 0 h h? 0 h sinx(0)(2) + cos(2x)(1)(2) h'(x) = 2cos(2x) The conjecture when the derivative of Sin(x+c) is Cos(x+c), is the same as in the original conjecture, in which the minimum and maximum slopes of Sinx determine the minimum and maximum values for Cosx. The only change is that the x0intercept change depending on the positive or negative value of c. An example of this can be seen below, in which sin(x + ?/4) and Cos (x + ?/4) are shown. j(x) = sin(x + 2) Lim j(x + h) - j(x) h? 0 h Lim sin((x + 2)+ h) - sin(x + 2) h? 0 h Lim sin(x + 2)cos(h) + sin(h)cos(x + 2) - sin(x + 2) h? 0 h Lim sin(x + 2)(cos(h) - 1) + sin(h)cos(x + 2) h? 0 h sin(x + 2) Lim (cos(h) - 1) + cos(x + 2) Lim sin(h) h? 0 h h? 0 h sin(x + 2)(0) + cos(x + 2)(1) h'(x) = cos(x + 2) 1 Note that this limit is present throughout all graphs in this paper, and that the said graphs increase and decrease in increments of . 2 Due to the fact that sin(x +h) = sin(x) multiplied by cos(h) + sin(h) multiplied by cos(x), I have taken the liberaty of substituting in the equation. ?? ?? ?? ?? 5

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