Derive formulas for the second and third derivatives

Derive formulas for the second and third derivatives of the inverse function $x = f^{-1}(y)$ of $y = f(x)$, expressing $\dfrac{d^{2}x}{dy^{2}}$ and $\dfrac{d^{3}x}{dy^{3}}$ through $y'_{x},\ y''_{xx},\ y'''_{xxx}$.

Step 1: The first derivative

Differentiating the identity $x = f^{-1}\bigl(f(x)\bigr)$ gives $x'_{y}\,y'_{x} = 1$, so

\[ x'_{y} = \frac{1}{y'_{x}}. \tag{1} \]

Step 2: The conversion rule

Everything below rests on one observation: for any function $\varphi$ of $x$, \[\frac{d\varphi}{dy} = \frac{d\varphi/dx}{dy/dx} = \frac{\varphi'_{x}}{y'_{x}}. \tag{$*$}\]

Differentiating with respect to $y$ therefore means "differentiate with respect to $x$, then divide by $y'_{x}$".

Step 3: The second derivative

Apply $(*)$ to (1): \[x''_{yy} = \frac{\left(\dfrac{1}{y'_{x}}\right)'_{x}}{y'_{x}} = \frac{-y''_{xx}/\bigl(y'_{x}\bigr)^{2}}{y'_{x}}.\]

\[x''_{yy} = -\dfrac{y''_{xx}}{\bigl(y'_{x}\bigr)^{3}}\]

Step 4: The third derivative

Apply $(*)$ once more. Writing $y',y'',y'''$ for the $x$-derivatives, the quotient rule gives

\[ \left(-\frac{y''}{\bigl(y'\bigr)^{3}}\right)'_{x} = -\frac{y'''\bigl(y'\bigr)^{3}-3\bigl(y'\bigr)^{2}\bigl(y''\bigr)^{2}}{\bigl(y'\bigr)^{6}} = \frac{3\bigl(y''\bigr)^{2}-y'''y'}{\bigl(y'\bigr)^{4}}, \]

and dividing by $y'$ once more: $x'''_{yyy} = \dfrac{3\bigl(y''_{xx}\bigr)^{2}-y'''_{xxx}\,y'_{x}}{\bigl(y'_{x}\bigr)^{5}}$.

Step 5: Verification with $y = x^{2}\ \ (x>0)$

Here $y' = 2x,\ y'' = 2,\ y''' = 0$, and the inverse is $x = y^{1/2}$.

\[x''_{yy} = -\frac{2}{\bigl(2x\bigr)^{3}} = -\frac{1}{4x^{3}}; \qquad \text{directly } -\tfrac14y^{-3/2} = -\frac{1}{4x^{3}} \;\;\checkmark\]

\[x'''_{yyy} = \frac{3\cdot4-0}{\bigl(2x\bigr)^{5}} = \frac{12}{32x^{5}} = \frac{3}{8x^{5}}; \qquad \text{directly } \tfrac38y^{-5/2} = \frac{3}{8x^{5}} \;\;\checkmark\]

\[\text{Note the pattern: order } n \text{ carries } \bigl(y'\bigr)^{-(2n-1)} \text{ in the denominator} \;\;\checkmark\]