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5128_Ch03_pp098-184

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Section 3.8 Derivatives of Inverse Trigonometric Functions 167<br />

EXAMPLE 1<br />

Applying the Formula<br />

d<br />

sin d x<br />

1 x 2 1 d<br />

1 x 2 <br />

<br />

2 • x d x<br />

2 2x<br />

<br />

1 x 4 <br />

Now try Exercise 3.<br />

Derivative of the Arctangent<br />

Although the function y sin 1 x has a rather narrow domain of 1, 1, the function<br />

y tan 1 x is defined for all real numbers, and is differentiable for all real numbers, as<br />

we will now see. The differentiation proceeds exactly as with the arcsine function.<br />

y tan 1 x<br />

tan y x<br />

d d<br />

tan y x<br />

d x d x<br />

sec 2 dy<br />

y 1 d x<br />

dy 1<br />

<br />

d x sec 2 y<br />

Inverse function relationship<br />

Implicit differentiation<br />

1<br />

1 tan y 2 Trig identity: sec 2 y 1 tan 2 y<br />

1<br />

<br />

1 x 2<br />

The derivative is defined for all real numbers. If u is a differentiable function of x, we get<br />

the Chain Rule form:<br />

d<br />

tan d x<br />

1 1<br />

u 1 u 2 d u<br />

.<br />

dx<br />

EXAMPLE 2 A Moving Particle<br />

A particle moves along the x-axis so that its position at any time t 0 is xt tan 1 t.<br />

What is the velocity of the particle when t 16?<br />

SOLUTION<br />

d<br />

vt tan d t<br />

1 1 d 1 1<br />

t • t • <br />

1 t 2 d t 1 t 2 t<br />

1 1 1<br />

When t 16, the velocity is v16 • Ṅow 1 16 2 16 1 36<br />

try Exercise 11.<br />

Derivative of the Arcsecant<br />

We find the derivative of y sec 1 x, x 1, beginning as we did with the other inverse<br />

trigonometric functions.<br />

y sec 1 x<br />

sec y x<br />

Inverse function relationship<br />

d d<br />

sec y x<br />

d x d x<br />

dy<br />

sec y tan y 1 d x<br />

dy 1<br />

d x sec y tan y<br />

Since ⏐x⏐ 1, y lies in (0, p2) (p2, p)<br />

and sec y tan y 0.

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