Question

Find the derivative of f(x) = sin^-1(x/2). Leave your answer in simplified form.

Answer 1

f'(x)= -1/2 [cos^(-3x/2)]
f'(x)=-1/[2 sec^(3x/2)]

Answer 2

\[y=\sin^{-1}(\frac{x}{2})=>\sin(y)=\frac{x}{2}\]
take derivative of both sides
\[\cos(y) y'=\frac{1}{2}=>y'=\frac{1}{2 \cos(y)}\]

but sin(y)=opp/hyp=x/2

using pythagorean thm we can find the adj side of y so we have
adj=sqrt{2^2-x^2}=sqrt{4-x^2}

so
\[y'=\frac{1}{2 \frac{\sqrt{4-x^2}}{2}}=\frac{1}{\sqrt{4-x^2}}\]