CALCULATING DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS

1. DERIVATIVES OF INVERSE
TRIGONOMETRIC FUNCTIONS

2. 𝑑
𝑑𝑥
𝑠𝑖𝑛−1
𝑥 =
1
1−𝑥2
𝑑
𝑑𝑥
𝑐𝑜𝑠−1
𝑥= -
1
1−𝑥2
𝑑
𝑑𝑥
𝑡𝑎𝑛−1 𝑥=
1
1+𝑥2
𝑑
𝑑𝑥
𝑐𝑜𝑡−1
𝑥=-
1
1+𝑥2
𝑑
𝑑𝑥
𝑠𝑒𝑐−1 𝑥 =
1
𝑥 𝑥2−1
𝑑
𝑑𝑥
𝑐𝑜𝑠𝑒𝑐−1
𝑥= -
1
𝑥 𝑥2−1

3. Question 1
Differentiate
𝑠𝑖𝑛−1
1
𝑥 + 1
w.r.t x
𝑑
𝑑𝑥
𝑠𝑖𝑛−1
1
𝑥 + 1
=
1
1 −
1
𝑥 + 1
𝑑
𝑑𝑥
1
𝑥 + 1
= −
1
𝑥 + 1 − 1
𝑥 + 1
1
2(𝑥 + 1)
3
2

4. = −
1
2
𝑥
𝑥 + 1
(𝑥 + 1) 𝑥 + 1
−
1
2 𝑥(𝑥 + 1)

5. Question 2
Differentiate
𝑠𝑖𝑛−1
𝑥 + 𝑠𝑖𝑛−1
1 − 𝑥2 w.r.t x
1
1 − 𝑥2
+
1
1 − (1 − 𝑥2)
𝑑
𝑑𝑥
1 − 𝑥2
1
1 − 𝑥2
+
1
2𝑥
(−2𝑥)
1 − 𝑥2
= 0

6. Question 3
Use substitution to find the derivative w.r.t x
𝑦 = 𝑡𝑎𝑛−1
1 − 𝑐𝑜𝑠𝑥
1 + 𝑐𝑜𝑠𝑥 cos 2𝑥 = 2 𝑐𝑜𝑠2
𝑥 − 1
= 1- 2 𝑠𝑖𝑛2 𝑥
𝑦 = 𝑡𝑎𝑛−1
2𝑠𝑖𝑛2 𝑥
2
2𝑐𝑜𝑠2 𝑥
2

7. = 𝑡𝑎𝑛−1
𝑡𝑎𝑛
𝑥
2
=
𝑥
2
𝑦 =
𝑥
2
𝑑𝑦
𝑑𝑥
=
1
2

8. Question 4
𝐼𝑓 𝑦 = 𝑠𝑖𝑛−1
(
2𝑥
1 + 𝑥2) Prove that (1+𝑥2
)
𝑑𝑦
𝑑𝑥
= 2
𝑝𝑢𝑡 𝑥 = tan 𝜃
𝑦 = si 𝑛−1
2𝑡𝑎𝑛𝜃
1 + 𝑡𝑎𝑛2 𝜃
= 𝑠𝑖𝑛−1(𝑠𝑖𝑛2𝜃) = 2𝜃

9. 𝑦 = 2𝜃 = 2𝑡𝑎𝑛−1 𝑥
𝑑𝑦
𝑑𝑥
=
2
1 + 𝑥2
1 + 𝑥2
𝑑𝑦
𝑑𝑥
= 2
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