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.