Find (d²x)/(dy²) and (d³x)/(dy³) for y = sin x

Find $\dfrac{d^{2}x}{dy^{2}}$ and $\dfrac{d^{3}x}{dy^{3}}$ for $y = \sin x$.

Step 1: Derivatives of $y = \sin x$

\[ y'_{x} = \cos x, \qquad y''_{xx} = -\sin x, \qquad y'''_{xxx} = -\cos x. \]

Step 2: Second derivative of the inverse

By the formula of Problem 324, $x''_{yy} = -\dfrac{y''_{xx}}{\bigl(y'_{x}\bigr)^{3}}$: $x''_{yy} = -\frac{-\sin x}{\cos^{3}x} = \frac{\sin x}{\cos^{3}x}$.

Since $y = \sin x$ and $\cos^{2}x = 1-y^{2}$ on $|x|\lt \tfrac{\pi}{2}$: $\frac{d^{2}x}{dy^{2}} = \dfrac{y}{\bigl(1-y^{2}\bigr)^{3/2}}$.

Step 3: Third derivative of the inverse

Now $x'''_{yyy} = \dfrac{3\bigl(y''_{xx}\bigr)^{2}-y'''_{xxx}y'_{x}}{\bigl(y'_{x}\bigr)^{5}}$, and the product of the two cosines contributes a plus sign: \[x'''_{yyy} = \frac{3\sin^{2}x-\bigl(-\cos x\bigr)\cos x}{\cos^{5}x} = \frac{3\sin^{2}x+\cos^{2}x}{\cos^{5}x} = \frac{2\sin^{2}x+1}{\cos^{5}x}.\]

\[\frac{d^{3}x}{dy^{3}} = \dfrac{2y^{2}+1}{\bigl(1-y^{2}\bigr)^{5/2}}\]

Step 4: Verification by differentiating the arcsine directly

The inverse is $x = \arcsin y$, with $x'_{y} = \bigl(1-y^{2}\bigr)^{-1/2}$.

\[x''_{yy} = -\tfrac12\bigl(1-y^{2}\bigr)^{-3/2}\bigl(-2y\bigr) = \frac{y}{\bigl(1-y^{2}\bigr)^{3/2}} \;\;\checkmark\]

\[x'''_{yyy} = \bigl(1-y^{2}\bigr)^{-3/2}+3y^{2}\bigl(1-y^{2}\bigr)^{-5/2} = \frac{\bigl(1-y^{2}\bigr)+3y^{2}}{\bigl(1-y^{2}\bigr)^{5/2}} = \frac{1+2y^{2}}{\bigl(1-y^{2}\bigr)^{5/2}} \;\;\checkmark\]

\[y = \tfrac12:\ x''_{yy} = \frac{0.5}{0.75^{3/2}} \approx 0.76980, \qquad x'''_{yyy} = \frac{1.5}{0.75^{5/2}} \approx 3.07920 \;\;\checkmark\]