Fig. 27-2
220
Fig. 27-1
Show that D^(sin x)
27.2
Let y = sin ' x. Then sin>' = x. By implicit differentiation, cos y • D x y = 1, D x y = I/cosy. But
Since, by definition,
—TT/2^y^ir/2,
cosysO, and, therefore.
and
CHAPTER 27
Inverse Trigonometric Functions
27.1
Draw the graph of y = sin ' x.
By definition, as x varies from -1 to l,y varies from -ir/2 to ir/2. The graph of y = sin"
J x is obtained
from the graph of y = sin x [Fig. 27-l(a)] by reflection in the line y = x. See Fig. 27-l(fc).
cos
cos
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220
Fig. 27-1
Show that D^(sin x)
27.2
Let y = sin ' x. Then sin>' = x. By implicit differentiation, cos y • D x y = 1, D x y = I/cosy. But
Since, by definition,
—TT/2^y^ir/2,
cosysO, and, therefore.
and
CHAPTER 27
Inverse Trigonometric Functions
27.1
Draw the graph of y = sin ' x.
By definition, as x varies from -1 to l,y varies from -ir/2 to ir/2. The graph of y = sin"
J x is obtained
from the graph of y = sin x [Fig. 27-l(a)] by reflection in the line y = x. See Fig. 27-l(fc).
cos
cos
Download
from Wow! eBook