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