Suppose f(x)=3sin^(−1)(x^4). Find: f'(x)

Answer 1

y = 3sin^-1(x^4)

y/3 = sin^-1(x^4)

sin(y/3) = sin[sin^-1(x^4)

sin(y/3) = x^4, cos(y/3) = sqr(1^2 - (x^4)^2) = sqr(1-x^8)

[sin(y/3)]' = 4x^3

cos(y/3)(1/3)y' = 4x^3

cos(y/3)y' = 12x^3

y' = 12x^3/cos(y/3) = 12x^3/sqr(1-x^8)

f'(x) = [(12x^3)(sqr(1-x^8)]/(1-x^8)

Answer 2

Two ways:

Method1:

y = 3sin^-1(x⁴)

sin(y/3) = x⁴

Take the implicit derivative of both sides

(cos(y/3)/3)y' = 4x³

Solve for y'

y' = 12x³/cos(y/3)

If sin(y/3) = x⁴ then cos(y/3) = √(1-x⁸)

f '(x) = 12x³/√(1-x⁸)

Method 2:

Use d(arcsin(z))/dx = (1/√(1-z²))dz/dx and z = x⁴

f'(x) = 3(1/√(1+(x⁴)²) (4x³) = 12x³/√(1-x⁸)