Find x'_y given y = sin x
Find $x'_{y}$ given $y = \sin x$.
Explanation of Method
Recovering the derivative of the arcsine from that of the sine.
Step 1: Differentiate the given relation
\[ y'_{x} = \cos x. \]
Step 2: Apply the inverse function rule
\[ x'_{y} = \frac{1}{y'_{x}} = \frac{1}{\cos x}, \qquad \cos x \neq 0. \]
Step 3: Express in terms of $y$
On the principal branch $-\tfrac{\pi}{2} \lt x \lt \tfrac{\pi}{2}$ the cosine is positive, so the sign of the square root is unambiguous: \[\cos x = \sqrt{1-\sin^{2}x} = \sqrt{1-y^{2}} \;\Longrightarrow\; x'_{y} = \frac{1}{\sqrt{1-y^{2}}}.\]
Step 4: Verification
\[\text{The inverse is } x = \arcsin y, \text{ whose standard derivative is } \frac{1}{\sqrt{1-y^{2}}} \;\;\checkmark\]
\[x = \frac{\pi}{6}:\quad y = \tfrac12,\quad x'_{y} = \frac{1}{\cos(\pi/6)} = \frac{2}{\sqrt3} \approx 1.15470 = \frac{1}{\sqrt{1-\tfrac14}} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,x'_{y} = \dfrac{1}{\cos x} = \dfrac{1}{\sqrt{1-y^{2}}}, \qquad |y| < 1\,}\]
Remark
Remark. This is how the derivative of $\arcsin$ is usually derived in the first place, rather than looked up. The restriction to a branch where $\cos x \gt 0$ is what forces the $+$ sign; on the branch $\tfrac{\pi}{2} \lt x \lt \tfrac{3\pi}{2}$ the same computation gives $-\dfrac{1}{\sqrt{1-y^{2}}}$, which is the derivative of that branch's inverse.