[1]

DIFFERENTIATION OF INVERSE

TRIGONOMETRIC FUNCTIONS

𝒇(𝒙)

𝒇′(𝒙)

𝒇(𝒙)

𝒇′(𝒙)

1. sin−1 𝑥 = 1

√1 − 𝑥2

sin−1 𝑢 = 1

√1 − 𝑢2 x

𝑑𝑢

𝑑𝑥

2. cos−1 𝑥 = −1

√1 − 𝑥2

cos−1 𝑢 = −1

√1 − 𝑢2 x

𝑑𝑢

𝑑𝑥

3. tan−1 𝑥 = 1

1 + 𝑥2

tan−1 𝑢 = 1

1 + 𝑢2 x

𝑑𝑢

𝑑𝑥

4. cot−1 𝑥 = −1

1 + 𝑥2

cot−1 𝑢 = −1

1 + 𝑢2 x 𝑑𝑢

𝑑𝑥

5. sec−1 𝑥 = 1

|𝑥|√𝑥2 − 1

sec−1 𝑢 = 1

|𝑢|√𝑢2 − 1 x

𝑑𝑢

𝑑𝑥

6. csc−1 𝑥 = −1

|𝑥|√𝑥2 − 1

csc−1 𝑢 = −1

|𝑢|√𝑢2 − 1 x

𝑑𝑢

𝑑𝑥

Example 1

Find dy

dx for 𝑦 = sin−1(4𝑥)

Solution: 𝐿𝑒𝑡 𝑢 = 4𝑥 𝑑𝑢

𝑑𝑥= 4

𝑦 = sin−1(𝑢)

𝑑𝑦

𝑑𝑥

= 1

√1 − 𝑢2 x

𝑑𝑢

𝑑𝑥

= 1

√1 − (4𝑥)2 x 4

= 𝟒

√𝟏 − 𝟏𝟔𝒙𝟐

[2]

Example 2

Find dy

dx for 𝑦 = tan−1(2𝑥 − 1)

Solution: 𝐿𝑒𝑡 𝑢 = (2𝑥 − 1) 𝑑𝑢

𝑑𝑥= 2

𝑦 = tan−1(𝑢)

𝑑𝑦

𝑑𝑥

= 1

1 + 𝑢2 x 𝑑𝑢

𝑑𝑥

= 1

1 + (2𝑥 − 1)2 x 2

= 2

1 + (2𝑥 − 1)2

= 2

1 + 4𝑥2 − 4𝑥 + 1

= 2

4𝑥2 − 4𝑥 + 2

= 2

2(2𝑥2 − 2𝑥 + 1)

= 𝟏

𝟐𝒙𝟐 − 𝟐𝒙 + 𝟏

[3]

Exercise 1

Determine 𝑑𝑦

𝑑𝑥 for the following equations:

1. 𝒚 = 𝐜𝐨𝐬−𝟏(𝟓𝒙)

𝐴𝑛𝑠𝑤𝑒𝑟 = −𝟓

√𝟏 − 𝟐𝟓𝒙𝟐

2. 𝒚 = 𝐭𝐚𝐧−𝟏(√𝒙𝟑)

𝐴𝑛𝑠𝑤𝑒𝑟 = 𝟑√𝒙

𝟐(𝟏 + 𝒙𝟑)

3. 𝒚 = 𝟑 𝐜𝐨𝐬−𝟏(𝒙𝟐 + 𝟎. 𝟓)

𝐴𝑛𝑠𝑤𝑒𝑟 = −𝟔𝒙

√𝒙𝟒 + 𝒙𝟐 + 𝟎. 𝟕𝟓

4. 𝒚 = 𝟒 𝐭𝐚𝐧−𝟏(𝟑𝒙𝟒)

𝐴𝑛𝑠𝑤𝑒𝑟 = 𝟒𝟖𝒙𝟑

𝟏 + 𝟗𝒙𝟖

5. 𝒚 = 𝟐√𝒙 𝐜𝐨𝐭−𝟏(𝒙)

𝐴𝑛𝑠𝑤𝑒𝑟 = 𝟐√𝒙 (−𝟏

𝟏 + 𝒙𝟐) + 𝐜𝐨𝐭−𝟏(𝒙) (𝟏

√𝒙)

6. 𝒚 = (𝒙𝟐 + 𝟏) 𝐬𝐢𝐧−𝟏(𝟒𝒙)

𝐴𝑛𝑠𝑤𝑒𝑟 =𝟒(𝒙𝟐 + 𝟏)

√𝟏 − 𝟏𝟔𝒙𝟐+ 𝟐𝒙 𝐬𝐢𝐧−𝟏(𝟒𝒙)

7. 𝒙 + 𝒚 = 𝐭𝐚𝐧−𝟏(𝒙𝟐 + 𝟑𝒚)

𝐴𝑛𝑠𝑤𝑒𝑟 =𝟐𝒙 − 𝟏 − (𝒙𝟐 + 𝟑𝒚)

𝟐

−𝟐 + (𝒙𝟐 + 𝟑𝒚)𝟐

8. 𝐬𝐢𝐧−𝟏(𝒙 + 𝒚) + 𝒚 = 𝒙𝟐

𝐴𝑛𝑠𝑤𝑒𝑟 =𝟐𝒙√𝟏 − (𝒙 + 𝒚)𝟐 − 𝟏

𝟏 + √𝟏 − (𝒙 + 𝒚)𝟐