University:
Massachusetts Institute of Technology
Course:
18.01 | Calculus
Academic year:

2010
Views:

283

Pages:

3
Author:

baluginii0vxt

Derivative of arctan(x) Let’s use our formula for the derivative of an inverse function to ﬁnd the deriva tive of the inverse of the tangent function: y = tan−1 x = arctan x. We simplify the equation by taking the tangent of both sides: y tan y = tan−1 x = tan(tan−1 x) tan y = x To get an idea what to expect, we start by graphing the tangent function (see Figure 1). The function tan(x) is deﬁned for − π2 < x < π2 . It’s graph extends from negative inﬁnity to positive inﬁnity. If we reﬂect the graph of tan x across the line y = x we get the graph of y = arctan x (Figure 2). Note that the function arctan x is deﬁned for all values of x from minus inﬁnity to inﬁnity, and limx→∞ tan−1 x = π2 . − π2 π 2 Figure 1: Graph of the tangent function. You may know that: d tan y dy = d sin y dy cos y .. . = = 1 1 cos2 y sec2 y Figure 2: Graph of tan−1 x. (If you haven’t seen this before, it’s good exercise to use the quotient rule to verify it!) We can now use implicit diﬀerentiation to take the derivative of both sides of our original equation to get: (Chain Rule) tan y d (tan(y)) dx d dy (tan(y)) dy dx � � 1 dy 2 cos (y) dx dy dx = = x d x dx = 1 = 1 = cos2 (y) Or equivalently, y � = cos2 y. Unfortunately, we want the derivative as a function of x, not of y. We must now plug in the original formula for y, which was y = tan−1 x, to get y � = cos2 (arctan(x)). This is a correct answer but it can be simpliﬁed tremendously. We’ll use some geometry to simplify it. y 1 (1+x2)1/2 x Figure 3: Triangle with angles and lengths corresponding to those in the exam ple. In this triangle, tan(y) = x so y = arctan(x). The Pythagorean theorem 2 tells us the length of the hypotenuse: h= � 1 + x2 and we can now compute: 1 cos(y) = √ . 1 + x2 From this, we get: 2 � cos (y) = so: 1 √ 1 + x2 �2 = 1 1 + x2 dy 1 = . dx 1 + x2 In other words, d 1 arctan(x) = . dx 1 + x2 3

Derivative of Arctan(x)

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