Inverse Trigonometric Functions Derivatives PDF

# Inverse Trigonometric Functions ## Derivatives | f(x) | d/dx f(x) | f(x) | d/dx f(x) | |:-----|:---------------|:----------------|:--------------------| | A | 0 | sin⁻¹(x) | 1 / √(1-x²) | | xⁿ | nxⁿ⁻¹ | cos⁻¹(x) | -1 / √(1-...

# Inverse Trigonometric Functions ## Derivatives | f(x) | d/dx f(x) | f(x) | d/dx f(x) | |:-----|:---------------|:----------------|:--------------------| | A | 0 | sin⁻¹(x) | 1 / √(1-x²) | | xⁿ | nxⁿ⁻¹ | cos⁻¹(x) | -1 / √(1-x²) | | eˣ | eˣ | tan⁻¹(x) | 1 / (1+x²) | | aˣ | aˣ ln(a) | cot⁻¹(x) | -1 / (1+x²) | | ln(x) | 1 / x | sec⁻¹(x) | 1 / (x√(x²-1)) | | Logₐ(x)| 1 / (x logₐ(e)) | csc⁻¹(x) | -1 / (x√(x²-1)) | | sin(x) | cos(x) | sinh⁻¹(x) | 1 / √(1+x²) | | cos(x) | -sin(x) | cosh⁻¹(x) | 1 / √(x²-1) | | tan(x) | sec²(x) | tanh⁻¹(x) | 1 / (1-x²) | | cot(x) | csc²(x) | coth⁻¹(x) | -1 / (x²-1) | | sec(x) | sec(x)tan(x) | sech⁻¹(x) | -1 / (x√(1-x²)) | | csc(x) | -csc(x)cot(x) | csch⁻¹(x) | -1 / (x√(1+x²)) | | sinh(x)| cosh(x) | coth(x) | -csch²(x) | | cosh(x)| sinh(x) | sech(x) | -sech(x)tanh(x) | | tanh(x)| sech²(x) | csch(x) | -csch(x)coth(x) | ## Proofs: ### 1. y = arcsin(x) = sin⁻¹(x) - Then: x = sin(y) - Derivative both sides with respect to x: - 1 = cos(y) * dy/dx - dy/dx = 1/cos(y) - Using the Pythagorean Identity: cos²(y) + sin²(y) = 1 - cos(y) = √(1-sin²(y)) = √(1-x²) - Therefore: dy/dx = 1 / √( 1-x² ) ### 2. y = arctan(x) = tan⁻¹(x) - Then: x = tan(y) - Derivative both sides with respect to x: - 1 = sec²(y) * dy/dx - dy/dx = 1/sec²(y) - Using the identity: 1 + tan²(y) = sec²(y) - dy/dx = 1 / (1 + tan²(y)) = 1/(1+x²) ### 3. y = arcsec(x) = sec⁻¹(x) - Then: x = sec(y) - Derivative both sides with respect to x: - 1 = sec(y)tan(y) * dy/dx - dy/dx = 1 / (sec(y)tan(y)) - Using the identity: 1 + tan²(y) = sec²(y) - tan(y) = √(sec²(y) - 1) = √(x²-1) - Therefore: dy/dx = 1 / (x√(x²-1))