Why a Second Derivative Needs Another Change of Variable
For a parametrized curve, both coordinates depend on a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). The first derivative \(\frac{dy}{dx}\) describes how \(y\) changes with respect to \(x\), even though the equations give each coordinate in terms of \(t\).
The second derivative \(\frac{d^2y}{dx^2}\) asks how that slope changes as \(x\) changes. A common source of error is to differentiate the slope with respect to \(t\) and stop. That produces the rate of change of slope with respect to the parameter, not with respect to \(x\). One more conversion is required.
Conditions and the Formula
Let \(x=x(t)\) and \(y=y(t)\) be twice differentiable on an open interval containing a parameter value \(t_0\). Suppose \(\frac{dx}{dt}\ne 0\) at \(t_0\). Then, in a sufficiently small interval around \(t_0\), \(x\) can be used as a local coordinate along the curve, and the parametric second derivative is given by:
The first derivative, which is needed inside this formula, is:
The twice-differentiability assumption ensures that the slope can be differentiated with respect to \(t\). The nonzero condition on \(\frac{dx}{dt}\) makes the change from \(t\) to \(x\) possible locally. Since \(\frac{dx}{dt}\) is continuous when \(x\) is twice differentiable, if it is nonzero at \(t_0\), it remains nonzero in some interval around \(t_0\).
Writing \(x'=\frac{dx}{dt}\) and \(y'=\frac{dy}{dt}\), the rule can also be expressed as:
The final equality follows by applying the quotient rule to \(\frac{y'}{x'}\). It is an equivalent form, not a separate rule with different conditions.
Deriving the Formula
The reason for dividing by \(\frac{dx}{dt}\) is the chain rule. In a small interval where \(x\) can be used as a coordinate, regard the slope \(\frac{dy}{dx}\) as a function of \(x\). That slope also varies with \(t\). The chain rule relates its rate of change with respect to \(t\) to its rate of change with respect to \(x\).
For a parametric curve, \(\frac{dy}{dx}=\frac{y'}{x'}\) when \(x'\ne 0\).
This gives \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), the rate at which the slope changes as the parameter changes.
The chain rule gives \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{dy}{dx}\right)\frac{dx}{dt}\).
Since \(\frac{dx}{dt}\ne 0\), divide by it to obtain \(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\).
For completeness, the local change of coordinate is justified by the nonzero derivative condition. Continuity of \(x'\) means that near \(t_0\), \(x'\) keeps the same sign as \(x'(t_0)\). By the Mean Value Theorem, \(x(t)\) is then strictly increasing or strictly decreasing on a sufficiently small interval, so it is one-to-one there. Its local inverse permits \(t\) to be viewed as a function of \(x\). The chain rule applies to the slope as a function of \(x\) and \(t\), giving the relationship in Step 3. Dividing is valid because \(x'\ne0\).
The expanded version follows directly. Starting from \(\frac{dy}{dx}=\frac{y'}{x'}\), the quotient rule gives:
Dividing this result by \(x'\) produces \(\frac{x'y''-y'x''}{(x')^3}\), as claimed. This derivation also shows why the denominator is the cube of \(x'\) in the expanded form.
Worked Examples
Worked Example: A Polynomial Parametrization
Suppose \(x=t^2+1\) and \(y=t^3+t\). Find \(\frac{d^2y}{dx^2}\) at \(t=1\), and verify it using the expanded formula.
First calculate the coordinate derivatives:
At \(t=1\), \(x'=2\ne0\), so the formula applies. The first derivative as a function of \(t\) is:
Differentiate this slope with respect to \(t\):
At \(t=1\), this derivative is \(\frac{3}{2}-\frac{1}{2}=1\). Divide by \(x'(1)=2\):
Check with the expanded formula. At \(t=1\), \(x'=2\), \(x''=2\), \(y'=4\), and \(y''=6\). Therefore:
Both forms give the same value, so the computation checks.
Worked Example: A Trigonometric Parametrization
Let \(x=\sin t\) and \(y=\cos t\). Find \(\frac{d^2y}{dx^2}\) at \(t=\frac{\pi}{4}\), checking the result against the curve written locally as \(y\) in terms of \(x\).
The derivatives are:
At \(t=\frac{\pi}{4}\), \(x'=\frac{\sqrt{2}}{2}\ne0\). The slope is:
Differentiate the slope with respect to \(t\), then divide by \(x'\):
Since \(\sec(\frac{\pi}{4})=\sqrt{2}\), the value is \(-(\sqrt{2})^3=-2\sqrt{2}\).
For a local check, the parametrization satisfies \(x^2+y^2=1\). Near \(t=\frac{\pi}{4}\), \(y\) is positive, so \(y=\sqrt{1-x^2}\). Differentiating gives \(dy/dx=-x(1-x^2)^{-1/2}\), and differentiating again gives \(d^2y/dx^2=-(1-x^2)^{-3/2}\). At \(x=\sqrt{2}/2\), this is \(-(\frac12)^{-3/2}=-2\sqrt{2}\), matching the parametric result.
Worked Example: A Linear \(x\)-Coordinate
Let \(x=3t-1\) and \(y=t^2+2t\). Find the second derivative at \(t=2\), and check it by expressing \(y\) in terms of \(x\).
Here \(x'=3\), \(x''=0\), \(y'=2t+2\), and \(y''=2\). In particular, \(x'=3\ne0\). The slope and its parameter derivative are:
Divide by \(x'=3\):
For a check, \(x=3t-1\) gives \(t=\frac{x+1}{3}\). Thus:
Differentiating twice with respect to \(x\) gives \(dy/dx=\frac{2x+8}{9}\) and \(d^2y/dx^2=\frac{2}{9}\), in agreement with the parametric calculation.
A Common Mistake: Forgetting the Conversion
Consider \(x=2t\) and \(y=t^2\), and find the second derivative at \(t=1\). The slope is:
A tempting but incorrect calculation is to differentiate this slope with respect to \(t\) and report \(1\) as the second derivative. That is only \(\frac{d}{dt}(\frac{dy}{dx})\). It has not yet been converted to a rate of change with respect to \(x\).
The correction is to divide by \(x'=2\):
This also checks by substitution: \(x=2t\) means \(t=x/2\), so \(y=t^2=x^2/4\), whose second derivative is \(\frac12\). The example shows the central distinction: differentiating with respect to \(t\) alone does not give the second derivative with respect to \(x\).
What to Check Before Applying the Rule
Before calculating a parametric second derivative, identify the parameter value and check the conditions there. The formula is not a license to divide by zero or to differentiate a slope that does not exist.
- Both coordinate functions should be twice differentiable near the parameter value for the displayed rule and expanded form.
- Check that \(\frac{dx}{dt}\ne0\) at the parameter value. This permits division and makes \(x\) a valid local coordinate.
- Form the first derivative \(\frac{dy}{dx}=\frac{y'}{x'}\) before differentiating it.
- After differentiating the slope with respect to \(t\), divide by \(x'\) once more.
If \(\frac{dx}{dt}=0\), the formula as stated cannot be used at that parameter value. That does not by itself settle whether a tangent or a second derivative can be described in some other way; it means this particular conversion requires a separate analysis.
Check Your Understanding
For each question, keep track of which variable is used for each differentiation.
- In one sentence, explain why \(\frac{d}{dt}(\frac{dy}{dx})\) is not generally equal to \(\frac{d^2y}{dx^2}\).
- State the parametric second derivative formula and the condition that permits its division.
- If \(x'=4\) and \(\frac{d}{dt}(\frac{dy}{dx})=-6\) at a parameter value, what is \(\frac{d^2y}{dx^2}\) there?
- For \(x=t^2\) and \(y=t^3\), find \(\frac{d^2y}{dx^2}\) at \(t=1\), showing the slope, its derivative with respect to \(t\), and the final division.
- What part of the standard formula fails if \(x'=0\) at the point under consideration?
