Derivatives of inverse trig functions
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This document discusses derivatives of inverse trigonometric functions and provides examples of taking their derivatives. It contains the following key points: 1. The derivatives of the main inverse trigonometric functions are given as fractions involving the term 1/(1-x^2). 2. Two examples are worked through, differentiating compositions of inverse trig functions and using trigonometric substitutions. 3. The last example proves that if y = sin^(-1)(2x/(1+x^2)), then (1+x^2)dy/dx = 2 by making a u-substitution of x = tanθ.
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Derivatives of inverse trig 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
- 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 **************************