Calculus 1

Derivatives of inverse trigonometric functions

Arcsine, arccosine and arctangent are built out of trigonometry. Why do their derivatives have no trigonometry left in them at all?

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Derivatives of inverse functionsCalculus 1 Inverse trig functionsPrecalculus

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Derivatives of inverse trigonometric functions

Why this exists

Odete stands in front of a tall billboard. Far away it looks small; right under the wall she sees it edge-on, and it looks small again. Somewhere between, the sign fills the widest angle of her view. That angle is a difference of two arctangents, and to say how fast it changes as she walks we need the derivative of arctangent.

The inverse trigonometric functions turn a ratio back into an angle, so they appear wherever an angle depends on lengths: a camera tracking a moving object, a trapdoor opening, a sight line to a sign. Their derivatives also matter in the other direction. The functions \dfrac{1}{1 + x^2} and \displaystyle \dfrac{1}{\sqrt{1 - x^2}} contain no trigonometry, yet they turn out to be the derivatives of arctangent and arcsine. That fact is what later lets us find areas under these algebraic curves.

Definition

Recall the slices chosen in Inverse trig functions:

Function	Means	Domain	Range
y = \arcsin x	\sin y = x	[-1, 1]	\displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]
y = \arccos x	\cos y = x	[-1, 1]	[0, \pi]
y = \arctan x	\tan y = x	all reals	\displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
y = \operatorname{arcsec} x	\sec y = x	\lvert x \rvert \ge 1	\displaystyle \left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]

The arcsecant row is a convention, and books differ on it. This lesson uses \operatorname{arcsec} x = \arccos\dfrac{1}{x}, which gives the range shown; some books instead take \displaystyle \left[0, \frac{\pi}{2}\right) \cup \left[\pi, \frac{3\pi}{2}\right), and their derivative formula looks different.

The derivative of arcsine at x is the slope of the graph of y = \arcsin x at the point (x, \arcsin x), and likewise for the others. This lesson finds each one as a formula in x.

Intuition

Figure 1The slope of y = \arcsin x is 1 at the origin, where the curve hugs its tangent y = x, \tfrac{5}{3} at x = 0.8, and grows without bound toward x = \pm 1, where the tangent is vertical.

The arcsine graph is the sine slice reflected in the line y = x. By the reciprocal-slope rule, the slope of arcsine at x is 1 divided by the slope of sine at the partner angle y = \arcsin x, which is \dfrac{1}{\cos y}.

Sine has slope 1 at 0, so arcsine does too. Toward the ends of the slice, sine flattens out: its slope \cos y shrinks to 0 at \displaystyle \pm\frac{\pi}{2}. Reflected, a flat stretch becomes a steep one, and the arcsine graph turns vertical at x = \pm 1.

What removes the trigonometry is a right triangle. If \sin y = x, draw a right triangle with hypotenuse 1 and side x opposite the angle y. The adjacent side is \displaystyle \sqrt{1 - x^2} by the Pythagorean theorem, so \displaystyle \cos y = \sqrt{1 - x^2}. The slope \dfrac{1}{\cos y} becomes \displaystyle \dfrac{1}{\sqrt{1 - x^2}}, a formula in x alone. At x = 0.8 the adjacent side is 0.6 and the slope is \tfrac{5}{3}, as in the figure.

Results

Derivatives of the inverse trigonometric functions.

\frac{d}{dx}\big[\arcsin x\big] = \frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1

\frac{d}{dx}\big[\arccos x\big] = -\frac{1}{\sqrt{1 - x^2}}, \quad -1 < x < 1

\frac{d}{dx}\big[\arctan x\big] = \frac{1}{1 + x^2}, \quad \text{all real } x

\frac{d}{dx}\big[\operatorname{arcsec} x\big] = \frac{1}{\lvert x \rvert\sqrt{x^2 - 1}}, \quad \lvert x \rvert > 1

Arcsine and arccosine are complementary. For every x in [-1, 1],

\arcsin x + \arccos x = \frac{\pi}{2}.

Endpoints. Arcsine and arccosine have no derivative at x = \pm 1; their graphs have vertical tangents there. Arctangent is differentiable everywhere.

With the chain rule. If u is a differentiable function of x, then wherever the right sides are defined,

\begin{aligned} \frac{d}{dx}\big[\arcsin u\big] &= \frac{u'}{\sqrt{1 - u^2}}, \\ \frac{d}{dx}\big[\arctan u\big] &= \frac{u'}{1 + u^2}. \end{aligned}

Why it is true

Arcsine by implicit differentiation. Let y = \arcsin x with -1 < x < 1. Then \sin y = x, and y lies strictly inside \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Differentiate both sides with respect to x, treating y as a function of x (Implicit differentiation):

\cos y \cdot \frac{dy}{dx} = 1, \qquad \frac{dy}{dx} = \frac{1}{\cos y}.

From \sin^2 y + \cos^2 y = 1 we get \cos^2 y = 1 - x^2, so \displaystyle \cos y = \pm\sqrt{1 - x^2}. The restricted range decides the sign: on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) cosine is positive, so \displaystyle \cos y = +\sqrt{1 - x^2}. Hence \displaystyle \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}. The positive sign matches the graph, which rises everywhere.

Arcsine by the reciprocal-slope rule. Take f(x) = \sin x on \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], so f^{-1} = \arcsin. The rule from Derivatives of inverse functions gives

\begin{aligned} \frac{d}{dx}\big[\arcsin x\big] &= \frac{1}{f'(\arcsin x)} \\ &= \frac{1}{\cos(\arcsin x)} \\ &= \frac{1}{\sqrt{1 - x^2}}. \end{aligned}

This route also settles a point the implicit route assumed: f'(y) = \cos y \neq 0 inside the slice, so the rule guarantees that arcsine is differentiable on (-1, 1). At \displaystyle y = \pm\frac{\pi}{2}, \cos y = 0, and the same lesson shows the inverse then has no derivative, with a vertical tangent.

Arctangent. Let y = \arctan x, so \tan y = x with y in \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Differentiating,

\begin{aligned} \sec^2 y \cdot \frac{dy}{dx} &= 1, \\ \frac{dy}{dx} = \frac{1}{\sec^2 y} &= \frac{1}{1 + \tan^2 y} \\ &= \frac{1}{1 + x^2}. \end{aligned}

No square root appears, so no sign has to be chosen. Since \sec^2 y \ge 1, tangent’s slope is never 0, and arctangent has a derivative at every real number.

Arccosine, and the minus sign. Let y = \arcsin x, in \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Then \displaystyle \frac{\pi}{2} - y lies in [0, \pi] and \displaystyle \cos\left(\frac{\pi}{2} - y\right) = \sin y = x, so \displaystyle \frac{\pi}{2} - y is the angle that arccosine picks: \displaystyle \arccos x = \frac{\pi}{2} - \arcsin x. Differentiating gives

\frac{d}{dx}\big[\arccos x\big] = 0 - \frac{1}{\sqrt{1 - x^2}}.

The minus sign says arccosine falls, from \pi at x = -1 to 0 at x = 1: its graph is the arcsine graph turned upside down and raised by \displaystyle \frac{\pi}{2}, so every slope changes sign. Exercise 15 reaches the same result directly from \cos y = x, and Exercise 16 derives the arcsecant formula.

With the chain rule. These follow from the formulas above and the chain rule, \dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}.

Worked examples

Example 1

Arcsine of a multiple

Let y = \arcsin(3x). Find \dfrac{dy}{dx}, say where it exists, and evaluate it at x = \tfrac{1}{6}.

Solution

The inside is u = 3x, with u' = 3. By the chain rule,

\frac{dy}{dx} = \frac{3}{\sqrt{1 - (3x)^2}} = \frac{3}{\sqrt{1 - 9x^2}}.

Arcsine needs -1 \le 3x \le 1, so the domain is \left[-\tfrac{1}{3}, \tfrac{1}{3}\right], and the derivative exists where 3x is strictly between -1 and 1, on \left(-\tfrac{1}{3}, \tfrac{1}{3}\right). At x = \pm\tfrac{1}{3} the graph has vertical tangents.

At x = \tfrac{1}{6}: 1 - 9 \cdot \tfrac{1}{36} = \tfrac{3}{4}, so

\frac{dy}{dx} = \frac{3}{\sqrt{3}/2} = \frac{6}{\sqrt{3}} = 2\sqrt{3}.

Check: at x = \tfrac{1}{6}, y = \arcsin\tfrac{1}{2} = \tfrac{\pi}{6}, and the reciprocal-slope rule with f(y) = \tfrac{1}{3}\sin y gives \displaystyle \dfrac{1}{\frac{1}{3}\cos\frac{\pi}{6}} = \dfrac{6}{\sqrt{3}}. The two agree.

Example 2

A tangent line to an arctangent

Find the tangent line to y = \arctan(2x) at x = \tfrac{1}{2}.

Solution

The point: y = \arctan 1 = \tfrac{\pi}{4}, since \displaystyle \tan\frac{\pi}{4} = 1 and \displaystyle \frac{\pi}{4} is in the range of arctangent.

The slope: with u = 2x, \dfrac{dy}{dx} = \dfrac{2}{1 + 4x^2}, which at x = \tfrac{1}{2} is \dfrac{2}{1 + 1} = 1.

The tangent line through \left(\tfrac{1}{2}, \tfrac{\pi}{4}\right) with slope 1 is

y = x - \frac{1}{2} + \frac{\pi}{4}.

Example 3

A derivative that collapses

Let \displaystyle F(x) = x\arccos x - \sqrt{1 - x^2} on (-1, 1). Find F'(x).

Solution

The first term is a product, so the product rule applies; the second is a composition, so the chain rule applies:

\begin{aligned} F'(x) &= \arccos x + x\left(-\frac{1}{\sqrt{1 - x^2}}\right) \\ &\qquad - \frac{-2x}{2\sqrt{1 - x^2}} \\ &= \arccos x - \frac{x}{\sqrt{1 - x^2}} + \frac{x}{\sqrt{1 - x^2}} \\ &= \arccos x. \end{aligned}

The two algebraic terms cancel, and what is left is arccosine itself. Read backward, F is a function whose derivative is \arccos x, which is exactly the kind of fact antiderivatives are built from.

Example 4

A corner and two vertical tangents

Let g(x) = \arcsin(1 - x^2). Find its domain and g'(x), and describe the graph at x = 0 and at the ends of the domain.

Solution

Arcsine needs -1 \le 1 - x^2 \le 1. The right inequality holds for every x; the left says x^2 \le 2. The domain is \displaystyle \left[-\sqrt{2}, \sqrt{2}\right].

With u = 1 - x^2 and u' = -2x,

g'(x) = \frac{-2x}{\sqrt{1 - (1 - x^2)^2}}.

Expand under the root: 1 - (1 - 2x^2 + x^4) = 2x^2 - x^4 = x^2(2 - x^2). Since \displaystyle \sqrt{x^2} = \lvert x \rvert,

g'(x) = \frac{-2x}{\lvert x \rvert\sqrt{2 - x^2}}

for \displaystyle 0 < \lvert x \rvert < \sqrt{2}.

So \displaystyle g'(x) = -\dfrac{2}{\sqrt{2 - x^2}} for x > 0 and \displaystyle g'(x) = \dfrac{2}{\sqrt{2 - x^2}} for x < 0.

At x = 0 the inside is u = 1, an endpoint of arcsine’s domain, and the formula gives \displaystyle \frac{0}{0}. The slopes approach \displaystyle -\sqrt{2} from the right and \displaystyle \sqrt{2} from the left, so the graph has a corner at \left(0, \tfrac{\pi}{2}\right) and no derivative there (Differentiability: corners, cusps and piecewise functions). The difference quotient agrees: for small h \neq 0, \dfrac{g(h) - g(0)}{h} is near \displaystyle -\sqrt{2} when h > 0 and near \displaystyle \sqrt{2} when h < 0.

At \displaystyle x = \pm\sqrt{2} the inside is u = -1, and \displaystyle \sqrt{2 - x^2} \to 0, so \lvert g'(x) \rvert grows without bound: vertical tangents at \displaystyle \left(\pm\sqrt{2}, -\tfrac{\pi}{2}\right). An endpoint of arcsine reached by the inside does not always give a vertical tangent; here the ends give vertical tangents and the middle gives a corner, because at x = 0 the inside 1 - x^2 has slope 0, and that zero meets the unbounded slope of arcsine at 1.

Example 5

Odete walks toward the billboard

Figure 2Example 5. The sign fills the angle \theta between the two sight lines, \theta = \arctan\dfrac{16}{x} - \arctan\dfrac{6}{x}.

The bottom edge of a billboard is 6 meters above Odete’s eye level and the top edge is 16 meters above it. She walks straight toward the wall at 1.5 meters per second. How fast is the angle \theta that the sign fills in her view changing when she is 12 meters from the wall?

Solution

Let x be her horizontal distance to the wall, in meters. The sight line to the top makes the angle \arctan\dfrac{16}{x} with the horizontal, and the one to the bottom makes \arctan\dfrac{6}{x}, so

\theta = \arctan\frac{16}{x} - \arctan\frac{6}{x}.

Differentiate with respect to x. For a constant c, the chain rule gives

\begin{aligned} \frac{d}{dx}\left[\arctan\frac{c}{x}\right] &= \frac{-c/x^2}{1 + c^2/x^2} \\ &= -\frac{c}{x^2 + c^2}. \end{aligned}

\frac{d\theta}{dx} = -\frac{16}{x^2 + 256} + \frac{6}{x^2 + 36}.

At x = 12: -\dfrac{16}{400} + \dfrac{6}{180} = -\dfrac{1}{25} + \dfrac{1}{30} = -\dfrac{1}{150} radian per meter.

She walks toward the wall, so x decreases: \dfrac{dx}{dt} = -1.5 meters per second. By the chain rule,

\frac{d\theta}{dt} = \frac{d\theta}{dx}\cdot\frac{dx}{dt} = \left(-\frac{1}{150}\right)(-1.5) = 0.01.

The angle grows at 0.01 radian per second, about 0.57^\circ per second. The sign of \dfrac{d\theta}{dx} changes where 6(x^2 + 256) = 16(x^2 + 36), that is at \displaystyle x = \sqrt{96} \approx 9.8 meters: closer than that, walking on makes the sign look smaller.

Boundaries

Arcsine and arccosine are differentiable only on the open interval (-1, 1). At x = \pm 1 the slope of sine or cosine at the partner angle is 0, and the graphs have vertical tangents. A composition \arcsin u can fail to be differentiable wherever u = \pm 1, as a vertical tangent or a corner (Example 4).
The formulas depend on the chosen ranges. The positive root in \displaystyle \dfrac{1}{\sqrt{1 - x^2}} comes from \cos y \ge 0 on \displaystyle \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. A different slice of sine would give a different sign (Exercise 22), and the arcsecant formula changes with the convention.
Angles are in radians. The formulas rest on \dfrac{d}{dx}\big[\sin x\big] = \cos x, which holds only in radians, and the inverse functions return radians.
\arctan(\tan x) = x only on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). Elsewhere a composition with the original function needs care (Exercise 25).

Where this goes next

Read backward, these formulas say that \dfrac{1}{1 + x^2} and \displaystyle \dfrac{1}{\sqrt{1 - x^2}} have arctangent and arcsine as antiderivatives, and Antiderivatives of trig, exponential and log functions uses exactly that. The same pattern, a slope of the original read at the partner point and then rewritten with an identity, gives the derivatives of logarithms and of any inverse a later course meets. In related rates, an angle given as an arctangent can now be differentiated directly, without first rewriting the relation with tangent.

Exercises

Derivatives from the formulas

Find each derivative and say where it exists.

y = \arcsin\dfrac{x}{2}
y = \arctan(5x)
y = \arccos(x^2)
\displaystyle y = \arctan\sqrt{x}
y = x^2\arctan x
y = \arcsin(e^x)
y = (\arctan x)^2
y = \arctan\dfrac{x - 1}{x + 1}
y = \ln(\arctan x)

Slopes and tangent lines

Find the slope of y = \arcsin x at x = \tfrac{1}{2} and at \displaystyle x = -\tfrac{\sqrt{3}}{2}.
Find the tangent line to y = \arctan x at \displaystyle x = \sqrt{3}.
Find the tangent line to y = \arccos x at x = 0.
At which points does y = \arctan x have slope \tfrac{1}{10}?
Let f(x) = \arcsin(x - 2). Find its domain, f'(x) and f'(2), and the points where the graph has vertical tangents.

Arccosine, arcsecant and identities

Derive the derivative of \arccos x by differentiating \cos y = x implicitly. Explain which square root to take and why.
Using \operatorname{arcsec} x = \arccos\dfrac{1}{x}, show that \displaystyle \dfrac{d}{dx}\big[\operatorname{arcsec} x\big] = \dfrac{1}{\lvert x \rvert\sqrt{x^2 - 1}} for \lvert x \rvert > 1, and find the slope at x = 2 and at x = -2.
Show that h(x) = \arctan x + \arctan\dfrac{1}{x} has h'(x) = 0 for every x \neq 0. Find h(1) and h(-1), and explain why h is not one constant on its whole domain.
Find \dfrac{d^2}{dx^2}\big[\arcsin x\big], and the interval on which the arcsine graph bends upward.

Rates

A trapdoor 1 meter long swings up about its hinge. Graca measures the height h of its far edge above the floor, in meters, so the angle with the floor is \theta = \arcsin h. The edge rises at 0.06 meters per second. How fast is \theta changing when h = 0.6?
Another sign has its bottom edge 3 meters and its top edge 12 meters above Graca’s eye level. Write the angle it fills when she stands x meters from the wall, show that \dfrac{d\theta}{dx} = 0 at x = 6, and find \theta there.
Joaquim films a jogger on a straight path from a camera 20 meters from the path. The camera turns to follow her, at angle \theta = \arctan\dfrac{x}{20} from the perpendicular, where x is her distance past the closest point. She runs at 8 meters per second. How fast does the camera turn when x = 15?

Reasoning

Suppose arcsine had been defined on the slice \displaystyle \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] of sine instead. What would its derivative be? Explain.
Explain why the graph of y = \arctan x has no vertical tangent, and why its slope approaches 0 as x \to \infty. How does this fit its horizontal asymptotes?
Arcsine is an odd function. Use the graph to explain why its derivative is an even function, and check this against the formula.
Show that \dfrac{d}{dx}\big[\arctan(\tan x)\big] = 1 wherever \tan x is defined, yet \arctan(\tan \pi) \neq \pi. Describe the graph of y = \arctan(\tan x).

Answers

\displaystyle \dfrac{1}{\sqrt{4 - x^2}} on (-2, 2). \displaystyle \dfrac{1/2}{\sqrt{1 - x^2/4}} = \dfrac{1/2}{\sqrt{4 - x^2}/2}.
\dfrac{5}{1 + 25x^2}, all real x.
\displaystyle -\dfrac{2x}{\sqrt{1 - x^4}} on (-1, 1). Chain rule with u = x^2, u' = 2x.
\displaystyle \dfrac{1}{2\sqrt{x}\,(1 + x)} for x > 0. \displaystyle u = \sqrt{x}, \displaystyle u' = \dfrac{1}{2\sqrt{x}}, 1 + u^2 = 1 + x.
2x\arctan x + \dfrac{x^2}{1 + x^2}, all real x, by the product rule.
\displaystyle \dfrac{e^x}{\sqrt{1 - e^{2x}}} for x < 0, where 0 < e^x < 1.
\dfrac{2\arctan x}{1 + x^2}, all real x. Outer function u^2, inner \arctan x.
\dfrac{1}{1 + x^2} for x \neq -1. u' = \dfrac{2}{(x + 1)^2} and 1 + u^2 = \dfrac{2x^2 + 2}{(x + 1)^2}, so \dfrac{u'}{1 + u^2} = \dfrac{2}{2x^2 + 2}.
\dfrac{1}{(1 + x^2)\arctan x} for x > 0, where \arctan x > 0.
\displaystyle \dfrac{2}{\sqrt{3}} = \dfrac{2\sqrt{3}}{3} and 2. 1 - \tfrac{1}{4} = \tfrac{3}{4}; 1 - \tfrac{3}{4} = \tfrac{1}{4} and \dfrac{1}{1/2} = 2.
\displaystyle y = \dfrac{\pi}{3} + \dfrac{1}{4}\left(x - \sqrt{3}\right). \displaystyle \arctan\sqrt{3} = \tfrac{\pi}{3}; slope \dfrac{1}{1 + 3}.
y = \dfrac{\pi}{2} - x. \arccos 0 = \tfrac{\pi}{2}; slope \displaystyle -\dfrac{1}{\sqrt{1}} = -1.
(3, \arctan 3) and (-3, -\arctan 3). \dfrac{1}{1 + x^2} = \dfrac{1}{10} gives x^2 = 9.
Domain [1, 3]; \displaystyle f'(x) = \dfrac{1}{\sqrt{1 - (x - 2)^2}} = \dfrac{1}{\sqrt{(x - 1)(3 - x)}} on (1, 3); f'(2) = 1; vertical tangents at \left(1, -\tfrac{\pi}{2}\right) and \left(3, \tfrac{\pi}{2}\right).
-\sin y \cdot \dfrac{dy}{dx} = 1, so \dfrac{dy}{dx} = -\dfrac{1}{\sin y}. \sin^2 y = 1 - x^2, and y is in (0, \pi) for -1 < x < 1, where sine is positive, so \displaystyle \sin y = \sqrt{1 - x^2} and \displaystyle \dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^2}}.
\displaystyle \dfrac{\sqrt{3}}{6} at both. The chain rule gives \displaystyle -\dfrac{1}{\sqrt{1 - 1/x^2}}\cdot\left(-\dfrac{1}{x^2}\right), which is \displaystyle \dfrac{1}{x^2\sqrt{(x^2 - 1)/x^2}}. Since \displaystyle \sqrt{x^2} = \lvert x \rvert, this is \displaystyle \dfrac{\lvert x \rvert}{x^2\sqrt{x^2 - 1}}, and \dfrac{\lvert x \rvert}{x^2} = \dfrac{1}{\lvert x \rvert}. At x = \pm 2: \displaystyle \dfrac{1}{2\sqrt{3}}. Both pieces of the graph rise.
h(1) = \dfrac{\pi}{2}, h(-1) = -\dfrac{\pi}{2}. \dfrac{d}{dx}\arctan\dfrac{1}{x} = \dfrac{-1/x^2}{1 + 1/x^2} = -\dfrac{1}{x^2 + 1}, which cancels \dfrac{1}{1 + x^2}. h(1) = 2\arctan 1. The domain is split at x = 0, and a zero derivative says nothing linking the two pieces. In fact for x > 0 the angles \arctan x and \displaystyle \arctan\frac{1}{x} are the two acute angles of a right triangle with legs 1 and x, so \displaystyle h = \frac{\pi}{2} there, and \displaystyle h = -\frac{\pi}{2} for x < 0 since h is odd.
\dfrac{x}{(1 - x^2)^{3/2}}; bends upward on (0, 1). Differentiate (1 - x^2)^{-1/2}: -\tfrac{1}{2}(1 - x^2)^{-3/2}(-2x). Positive exactly when x > 0.
\dfrac{3}{40} = 0.075 radian per second. \displaystyle \dfrac{d\theta}{dh} = \dfrac{1}{\sqrt{1 - 0.36}} = \dfrac{5}{4}; times 0.06.
\theta = \arctan\dfrac{12}{x} - \arctan\dfrac{3}{x}, \dfrac{d\theta}{dx} = -\dfrac{12}{x^2 + 144} + \dfrac{3}{x^2 + 9}, which at 6 is -\dfrac{12}{180} + \dfrac{3}{45} = 0. There \theta = \arctan 2 - \arctan\tfrac{1}{2} = \arctan\tfrac{3}{4} \approx 0.644 radian, by the tangent subtraction formula: \dfrac{2 - 1/2}{1 + 1} = \dfrac{3}{4}.
\dfrac{32}{125} = 0.256 radian per second. \dfrac{d\theta}{dx} = \dfrac{1/20}{1 + (15/20)^2} = \dfrac{1}{20}\cdot\dfrac{16}{25} = \dfrac{4}{125}; times 8.
\displaystyle -\dfrac{1}{\sqrt{1 - x^2}}. The implicit step still gives \dfrac{1}{\cos y}, but on \displaystyle \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) cosine is negative, so \displaystyle \cos y = -\sqrt{1 - x^2}. Sine falls on that slice, so its inverse falls too.
\dfrac{1}{1 + x^2} is defined for every x, so the slope is a finite number everywhere: tangent’s slope \sec^2 y is never 0. As x \to \infty, 1 + x^2 \to \infty, so the slope tends to 0, matching a graph that levels off toward its asymptote \displaystyle y = \frac{\pi}{2} (and toward \displaystyle y = -\frac{\pi}{2} as x \to -\infty).
The graph of an odd function is unchanged by a half-turn about the origin, and a half-turn carries a tangent line to a parallel one, so the slopes at x and -x agree. The formula \displaystyle \dfrac{1}{\sqrt{1 - x^2}} depends only on x^2.
By the chain rule, \dfrac{\sec^2 x}{1 + \tan^2 x} = 1. But \tan\pi = 0 and \arctan 0 = 0, not \pi, because arctangent returns an angle in \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right). The graph is a sawtooth: segments of slope 1, each running from height \displaystyle -\frac{\pi}{2} to \displaystyle \frac{\pi}{2}, with breaks at the odd multiples of \displaystyle \frac{\pi}{2}, where \tan x is undefined. On the piece through x = \pi it is y = x - \pi.