Understanding the Second Derivative Relationship

The question asks for the second derivative of \(x\) with respect to \(y\), which is denoted as \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\). We need to express this in terms of the derivatives of \(y\) with respect to \(x\), namely \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) and \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}}\).

First Derivative of x with respect to y

We know the relationship between the first derivatives:

\[\frac{{{\rm{dx}}}}{{{\rm{dy}}}} = \frac{1}{{\frac{{{\rm{dy}}}}{{{\rm{dx}}}}}}\]

We can write this using negative exponents as:

\[\frac{{{\rm{dx}}}}{{{\rm{dy}}}} = {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 1}}\]

Calculating the Second Derivative of x with respect to y

Now, to find the second derivative \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\), we need to differentiate \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}}\) with respect to \(y\). Using the chain rule, differentiating with respect to \(y\) is equivalent to differentiating with respect to \(x\) and then multiplying by \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}}\).

So, we have:

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = \frac{{\rm{d}}}{{{\rm{dy}}}}\left( {\frac{{{\rm{dx}}}}{{{\rm{dy}}}}} \right) = \frac{{\rm{d}}}{{{\rm{dy}}}}\left( {{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 1}}} \right)\]

Applying the chain rule \(\frac{{{\rm{d}}f}}{{{\rm{dy}}}} = \frac{{{\rm{d}}f}}{{{\rm{dx}}}} \cdot \frac{{{\rm{dx}}}}{{{\rm{dy}}}}\):

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 1}}} \right) \cdot \frac{{{\rm{dx}}}}{{{\rm{dy}}}}\]

Let's first calculate \(\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 1}}} \right)\) using the power rule and chain rule:

\[\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 1}}} \right) = - 1 \cdot {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 1 - 1}} \cdot \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)\]

\[ = - {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 2}} \cdot \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}}\]

Now substitute this back into the expression for \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\):

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = \left[ { - {{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)}^{ - 2}} \cdot \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}}} \right] \cdot \frac{{{\rm{dx}}}}{{{\rm{dy}}}}\]

We know that \(\frac{{{\rm{dx}}}}{{{\rm{dy}}}} = {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 1}}\). Substitute this in:

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = - {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 2}} \cdot \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}} \cdot {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 1}}\]

Combine the terms with \(\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)\):

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = - \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}} \cdot {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 2 - 1}}\]

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = - \frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}} \cdot {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 3}}\]

Rearranging the terms, we get:

\[\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}} = - \left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}} \right){\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 3}}\]

Comparison with Options

Let's compare our derived expression with the given options:

• Option 1: \( - {\left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{x}}^2}}}} \right)^{ - 1}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 3}}\) - Does not match
• Option 2: \({\left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}}} \right)^{ - 1}}{\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 2}}\) - Does not match
• Option 3: \( - \left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}}} \right){\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 3}}\) - Matches our result
• Option 4: \({\left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}}} \right)^{ - 1}}\) - Does not match

The calculated expression for \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\) is \( - \left( {\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}} \right){\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 3}}\).

Revision Table: Key Differentiation Concepts

Additional Information on Higher-Order Derivatives

Calculating higher-order derivatives of inverse functions requires careful application of the chain rule repeatedly. The formula for \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\) is a standard result derived using these fundamental rules.

Understanding how to switch the variable of differentiation (from \(x\) to \(y\) or vice versa) using the chain rule \(\frac{{\rm{d}}}{{{\rm{dy}}}} = \frac{{\rm{d}}}{{{\rm{dx}}}} \cdot \frac{{{\rm{dx}}}}{{{\rm{dy}}}}\) is crucial. This allows us to express derivatives with respect to \(y\) in terms of derivatives with respect to \(x\).

This problem demonstrates the process of finding \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\) in terms of \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\) and \(\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{dx}}^2}}}\). Higher-order derivatives like \(\frac{{{{\rm{d}}^3}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^3}}}\) would follow a similar but more complex process, involving further differentiation of the expression for \(\frac{{{{\rm{d}}^2}{\rm{x}}}}{{{\rm{d}}{{\rm{y}}^2}}}\) with respect to \(y\).