Asked by Jim
find the derivative of the function. Simplify where possible.
F(theta)=arcsin(square root of (sin9(theta)))
F(theta)=arcsin(square root of (sin9(theta)))
Answers
Answered by
Steve
f(θ) = arcsin(√sin9θ)
since d/dx arcsin(x) = 1/√(1-x^2),
f'(θ) = 1/√(1-sin9θ) * 9cos9θ/2√sin9θ
= 9cos9θ / 2√(sin9θ)√(1-sin9θ)
nasty, but that's how it is
since d/dx arcsin(x) = 1/√(1-x^2),
f'(θ) = 1/√(1-sin9θ) * 9cos9θ/2√sin9θ
= 9cos9θ / 2√(sin9θ)√(1-sin9θ)
nasty, but that's how it is
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