3.8 Derivatives of Inverse Trigonometric Functions

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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:

Derivative Formula for Inverses:

df

dx dfdx

x f a

x a

1 1

( )

evaluated at ( )f a

is equal to the reciprocal of

the derivative of ( )f x

evaluated at .a

The derivative of 1( )f x

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A typical problem using this formula might look like this:

Given: 3 5f 3 6df

dx

Find: 1

5df

dx

Derivative Formula for Inverses:

df

dx dfdx

x f a

x a

1 1

( )

1 1

56

df

dx

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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

Your calculator contains all six inverse trig functions.However it is occasionally still useful to know the following:

1 1 1sec cosx

x

1 1cot tan2

x x

1 1 1csc sinx

x

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