DERIVATIVES OF INVERSE TRIG FUNCTIONS

𝑦=arc𝑠𝑖𝑛𝑥 𝑦=𝑠𝑖𝑛− 1𝑥

sin 𝑦=𝑥

𝑑𝑑𝑥sin 𝑦=

𝑑𝑑𝑥

𝑥

cos 𝑦𝑑𝑦𝑑𝑥

=1

𝑑𝑦𝑑𝑥

=1

cos𝑦=

1

√1−𝑠𝑖𝑛2 𝑦𝑑𝑑𝑥

𝑎𝑟𝑐 sin 𝑥=1

√1−𝑥2

𝑦=arc𝑐𝑜𝑠𝑥 𝑦=𝑐𝑜𝑠− 1𝑥

cos 𝑦=𝑥𝑑𝑑𝑥cos𝑦=

𝑑𝑑𝑥

𝑥

−sin 𝑦𝑑𝑦𝑑𝑥

=1

𝑑𝑦𝑑𝑥

=1

−𝑠𝑖𝑛 𝑦=−

1

√1−𝑐𝑜𝑠2 𝑦𝑑𝑑𝑥

𝑎𝑟𝑐cos𝑥=−1

√1−𝑥2

𝑦=arc 𝑡𝑎𝑛𝑥𝑦=𝑡𝑎𝑛− 1𝑥

𝑡𝑎𝑛 𝑦=𝑥

𝑑𝑑𝑥

𝑡𝑎𝑛 𝑦=𝑑𝑑𝑥

𝑥

𝑐𝑜𝑠2 𝑦+𝑠𝑖𝑛2 𝑦𝑐𝑜𝑠2𝑦

𝑑𝑦𝑑𝑥

=1

𝑑𝑦𝑑𝑥

=𝑐𝑜𝑠2 𝑦= 𝑐𝑜𝑠2 𝑦𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2 𝑦

𝑑𝑑𝑥

𝑎𝑟𝑐 tan𝑥=1

1+𝑥2

𝑦=arc𝑐𝑜𝑡 𝑥 𝑦=𝑐𝑜𝑡− 1𝑥

𝑐𝑜𝑡 𝑦=𝑥

𝑑𝑑𝑥co𝑡 𝑦=

𝑑𝑑𝑥

𝑥

−𝑠𝑖𝑛2 𝑦−𝑐𝑜𝑠2 𝑦𝑠𝑖𝑛2 𝑦

𝑑𝑦𝑑𝑥

=1

𝑑𝑑𝑥

𝑎𝑟𝑐cot 𝑥=−1

1+𝑥2

𝑑𝑦𝑑𝑥

=𝑐𝑜𝑠2 𝑦 ∙

1

𝑐𝑜𝑠2 𝑦

(𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2 𝑦 ) ∙ 1𝑐𝑜𝑠2 𝑦

𝑑𝑦𝑑𝑥

=−𝑠𝑖𝑛2 𝑦=− 𝑠𝑖𝑛2 𝑦𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2𝑦

𝑑𝑦𝑑𝑥

=−𝑠𝑖𝑛2 𝑦 ∙

1

𝑠𝑖𝑛2𝑦

(𝑠𝑖𝑛2 𝑦+𝑐𝑜𝑠2 𝑦 ) ∙ 1𝑠𝑖𝑛2 𝑦

To get derivatives of inverse trigonometric functions we were able to use implicit

differentiation.

Sometimes it is not possible/plausible to explicitly find inverse function, but we still want to find derivative of inverse function at certain point (slope).

QUESTION: What is the relationship between derivatives of a function and its inverse ????

Let and be functions that are differentiable everywhere.If is the inverse of and if and , what is ?

DERIVATIVE OF THE INVERSE FUNCTIONSexample:

Since is the inverse of you know that

holds for all .

Differentiating both sides with respect to , and using the the chain rule:

.

So ⇒ ⇒

The relation

𝑔 ′ (𝑥 )= 1𝑓 ′ (𝑔 (𝑥 ))

( 𝑓 − 1)′

(𝑥 )= 1𝑓 ′ ( 𝑓 − 1 (𝑥 ))

used here holds whenever and are inverse functions. Some people memorize it. It is easier to re-derive it any time you want to use it, by differentiating both sides of

(which you should know in the middle of the night).

A typical problem using this formula might look like this:

Given: 3 5f 3 6df

dx Find:

1

5df

dx

1 15

6

df

dx

example:

𝑓 (𝑔(𝑥))=𝑥

𝑓 ′ (𝑔(𝑥))𝑔 ′ (𝑥)=1

𝑓 (3 )=5⇒𝑔 (5 )=3

.

.

If , find

example:

𝑓 (𝑔(𝑥))=𝑥

𝑓 ′ (𝑔(𝑥))𝑔 ′ (𝑥)=1 .

If , then

•f -1(a) = b.•(f -1)’(a) = tan .

• f’(b) = tan

• + = π/2

( 𝑓 − 1 )′=tan=tan (𝜋2 −𝜃)¿cot 𝜃=

1tan 𝜃

=1

𝑓 ′ (𝑏)

( 𝑓 − 1 )′ (𝑎)=1

𝑓 ′ ( 𝑓 −1(𝑎)) 𝑡𝑟𝑢𝑒 𝑓𝑜𝑟 𝑎𝑛𝑦 𝑎 , 𝑠𝑜:( 𝑓− 1 )′ (𝑥)=

1

𝑓 ′ ( 𝑓 −1(𝑥))

Graphical Interpretation

Derivative of the inverse function at a point is the reciprocal of the derivative of the function at the corresponding point.

Slope of the line tangent to at is the reciprocal of the slope of at .

example:

If f(3) = 8, and f’(3)= 5, what do we know about f-1 ?

Since f passes through the point (3, 8), f-1 must pass through the point (8, 3). Furthermore, since the graph of f has slope 5 at (3, 8), the graph of f-1 must have slope 1/5 at (8, 3).

If f (x)= 2x5 + x3 + 1, find (a) f (1) and f '(1) and (b) (f -1 )(4) and (f -1)'(4).

y=2x5 + x3 + 1. y’ = 10x4 + 3x2 is positive everywhere y is strictly increasing, thus f (x) has an inverse.

example:

f (1)=2(1)5 + (1)3 +1=4f '(x)=10x4 + 3x2

f '(1)=10(1)4 +3(1)2 =13

Since f(1)=4 implies the point (1, 4) is on the curve f(x)=2x5 +x3 +1, therefore, the point (4, 1) (which is the reflection of (1, 4) on y =x) is on the curve (f -1)(x). Thus, (f -1)(4)=1.

example:

If f(x)=5x3 + x +8, find (f -1) '(8). Since y is strictly increasing near x =8, f(x) has an inverse near x =8.Note that f(0)=5(0)3 +0+8=8 which implies the point (0, 8) is on the curve of f(x). Thus, the point (8, 0) is on the curve of (f -1)(x).

f '(x)=15x2 +1f '(0)=1

http://www.millersville.edu/~bikenaga/calculus/inverse-functions/inverse-functions.html

We have been more careful than usual in our statement of the differentiability result for inverse functions. You should notice that the differentiation formula for the inverse function involves division by f '(f -1(x)). We must therefore assume that this value is not equal to zero. There is also a graphical explanation for this necessity.

Example. The graphs of the cubing function f(x) = x3 and its inverse (the cube root function) are shown below.

Notice that f '(x)=3x2 and so f '(0)=0. The cubing function has a horizontal tangent line at the origin. Taking cube roots we find that f -1(0)=0 and so f '(f -1(0))=0. The differentiation formula for f -1 can not be applied to the inverse of the cubing function at 0 since we can not divide by zero. This failure shows up graphically in the fact that the graph of the cube root function has a vertical tangent line (slope undefined) at the origin.

h

afhafaf

h

)()()( lim

0

Recognizing a given limit as a derivative (!!!!!!)

Example:

Example:

Example:

Tricky, isn’t it? A lot of grey cells needed.