Warmup: 1)
Warmup:
1)
3.8: Derivatives of Inverse Trig Functions
2 0f x x x
We can find the inverse function as follows:
2y x Switch x and y.2x y
x y
y x
2y x
y x
2df
xdx
At x = 2:
22 2 4f
2 2 2 4df
dx
4m 2,4
1f x x
1
1 2f x x 112
1
2
dfx
dx
1 1
2
df
dx x
To find the derivative of the inverse function:
2 0f x x x 2y x
y x
2df
xdx
At x = 2:
22 2 4f
2 2 2 4df
dx
4m 2,4
1f x x
1 1
2
df
dx x
1 1 1 14
2 2 42 4
df
dx
At x = 4:
1 4 4 2f
4,21
4m
Slopes are reciprocals.
2y x
y x
4m 2,4
4,21
4m
Slopes are reciprocals.
Because x and y are reversed to find the reciprocal function, the following pattern always holds:
evaluated at ( )f a
is equal to the reciprocal of
the derivative of ( )f x
evaluated at .a
The derivative of 1( )f x
The Rule for Inverses:Let f be a function that is differentiable on an interval. If f hasan inverse function g, the g is differentiable at any x for which
))((
1)(
xgfxg
In other words if )(
1)(
dx
d ),()(
1
11
xffxfthenxfxg
A typical problem using this formula might look like this:
)5(f Find
6(3)f 5 f(3) :
1- Given
6
1
)3(
1
))5((
1
))5((
11
fgfor
ff
** if f(3)=5, then g(5)=3
(3)?g of value theisWhat
x.allfor )(fg(x) and abledifferenti
is gfunction The -2.(6)f and -8,(3)f 3,f(6)
15,f(3)such that function abledifferenti a be f
1-
x
Let
)()( 1 xfxg
))((
1)3()(
1
1
xfff
since f(6) = 3 , 6)3(1 f
)6(
1)3()( 1
ff
2
1)3()( 1
f
x.allfor )(fg(x) and abledifferenti is gfunction
theand 52 x f(x) given that ),5(g 1-
3
x
xFind
Ans:
Since we do not know g(5) which we need to remember thatit is an inverse of f, so if g(5) = a, then f(a) = 5.
))((
1)(
xgfxg
set
0g(5) means which 5f(0)or 0, x
0)2(
02
552)(
2
3
3
so
xx
xx
xxxf
2
1
23(0)
1
)0(
1
))5((
1)(
2
f
gfxg
siny x
1siny xWe can use implicit differentiation to find:
1sind
xdx
1siny x
sin y x
sind d
y xdx dx
cos 1dy
ydx
1
cos
dy
dx y
We can use implicit differentiation to find:
1sind
xdx
1siny x
sin y x
sind d
y xdx dx
cos 1dy
ydx
1
cos
dy
dx y
2 2sin cos 1y y 2 2cos 1 siny y
2cos 1 siny y
But2 2
y
so is positive.cos y
2cos 1 siny y
2
1
1 sin
dy
dx y
2
1
1
dy
dx x
We could use the same technique to find and
.
1tand
xdx
1secd
xdx
1
2
1sin
1
d duu
dx dxu
12
1tan
1
d duu
dx u dx
1
2
1sec
1
d duu
dx dxu u
1
2
1cos
1
d duu
dx dxu
12
1cot
1
d duu
dx u dx
1
2
1csc
1
d duu
dx dxu u
1 1cos sin2
x x 1 1cot tan2
x x 1 1csc sec2
x x
Remember arcsin x as written be alsocan sin 1 x
2-1 xsin f(x)for )( xfFind
4-1 5x sec f(x)for )( xfFind
the end