Deriving the Second Derivative Formula

Why a Second Derivative Needs Another Chain Rule

A parametric curve is described by two functions of a parameter: \(x=x(t)\) and \(y=y(t)\). The first derivative \(dy/dx\) measures how \(y\) changes as \(x\) changes, even though both coordinates are initially given in terms of \(t\).

To obtain the second derivative, we need to differentiate that slope with respect to \(x\), not with respect to \(t\). The chain rule explains how to make this change of independent variable. The essential condition is that \(dx/dt\) is nonzero, so that \(x\) can serve locally as the independent variable.

The parametric slope formula is \(dy/dx=(dy/dt)/(dx/dt)\) when \(dx/dt\ne 0\). We will apply the chain rule to this slope and identify precisely when the result is valid.

Key idea: Differentiation with respect to \(x\) is related to differentiation with respect to \(t\) by \( \frac{d}{dx}=\frac{1}{dx/dt}\frac{d}{dt} \), provided \(dx/dt\ne 0\) and \(x\) is a valid local coordinate along the curve.

Conditions for Using \(x\) as the Independent Variable

Let \(x(t)\) and \(y(t)\) be twice continuously differentiable on an open interval containing \(t_0\). Suppose \(x'(t_0)\ne 0\). By continuity, \(x'\) remains nonzero and has a constant sign on a sufficiently small interval around \(t_0\). The mean value theorem then implies that \(x(t)\) is strictly monotone on that interval, so it has a local inverse \(t=t(x)\).

The curve can therefore be written locally as the graph \(y=Y(x)\), where \(Y(x)=y(t(x))\). Differentiating \(x(t(x))=x\) gives \(x'(t)\,dt/dx=1\), and hence \(dt/dx=1/x'(t)\). The chain rule applied to \(Y(x)=y(t(x))\) gives \(dY/dx=y'(t)/x'(t)\).

Condition to remember: The second-derivative formula below applies at parameter values where \(x'(t)\ne 0\), with \(x\) a valid local independent variable. A regular parameter value, where the velocity pair \((x'(t),y'(t))\) is not \((0,0)\), need not satisfy this condition: it is possible to have \(x'(t)=0\) but \(y'(t)\ne 0\).

Deriving the Formula

Write the parametric slope as \(p(t)=dy/dx\). Since \(p\) is expressed in terms of \(t\), the chain rule gives \(dp/dx=(dp/dt)(dt/dx)\). The inverse relation above supplies \(dt/dx=1/(dx/dt)\). Thus the second derivative is the derivative of the parametric slope with respect to \(t\), divided by \(dx/dt\).

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}, \qquad \frac{dx}{dt}\ne 0.

This is a chain-rule identity, not a new definition of the second derivative. Its meaning is the ordinary derivative of the local graph’s slope with respect to \(x\), expressed using the parameter \(t\). The denominator is required because changing \(t\) and changing \(x\) are not generally the same operation.

The quotient rule gives an equivalent expression. From \(dy/dx=y'/x'\), differentiate with respect to \(t\), then divide by \(x'\):

\frac{d}{dt}\left(\frac{y'}{x'}\right) = \frac{x'y''-y'x''}{(x')^2}, \qquad \frac{d^2y}{dx^2} = \frac{x'y''-y'x''}{(x')^3}, \qquad x'\ne 0.

Here primes denote derivatives with respect to \(t\). The two forms are equal because the quotient rule computes the \(t\)-derivative of the slope, while the final division by \(x'\) converts that rate of change into a rate of change with respect to \(x\).

Find the slope.
Compute \(dy/dx=y'/x'\), requiring \(x'\ne 0\).

Differentiate the slope in \(t\).
Apply \(d/dt\) to the entire expression \(y'/x'\).

Convert the rate to an \(x\)-rate.
Divide the result by \(x'\), again requiring \(x'\ne 0\).

Worked Examples: Seeing the Chain Rule at Work

Worked Example: A Linear \(x(t)\)

Let \(x(t)=3t-2\) and \(y(t)=t^2+1\). Find the second derivative at \(t=1\) using the parametric formula, then check it by writing \(y\) directly as a function of \(x\).

First differentiate both coordinates:

x'(t)=3,\qquad y'(t)=2t.

The slope is

\frac{dy}{dx}=\frac{2t}{3}.

Differentiate this slope with respect to \(t\), and divide by \(x'(t)\):

\frac{d}{dt}\left(\frac{2t}{3}\right)=\frac{2}{3}, \qquad \frac{d^2y}{dx^2} = \frac{2/3}{3} = \frac{2}{9}.

For a check, solve \(x=3t-2\) to obtain \(t=(x+2)/3\). Then \(y=(x+2)^2/9+1\), so \(dy/dx=2(x+2)/9\) and \(d^2y/dx^2=2/9\). This agrees with the parametric calculation, including at \(t=1\), where \(x=1\).

Worked Example: A Nonlinear \(x(t)\)

Let \(x(t)=t^2\) and \(y(t)=t^3\), and evaluate the second derivative at \(t=2\). Since \(x'(t)=2t\), the required condition holds near \(t=2\).

The coordinate derivatives are

x'(t)=2t,\qquad y'(t)=3t^2.

Therefore the slope, as a function of \(t\), is

\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2} \qquad (t\ne 0).

Differentiate this expression and divide by \(dx/dt\):

\frac{d}{dt}\left(\frac{3t}{2}\right)=\frac{3}{2}, \qquad \frac{d^2y}{dx^2} = \frac{3/2}{2t} = \frac{3}{4t}.

At \(t=2\), the value is \(3/(4\cdot2)=3/8\). To check, near \(t=2\) we have \(t>0\), so \(y=x^{3/2}\). Differentiating gives \(dy/dx=(3/2)x^{1/2}\) and \(d^2y/dx^2=(3/4)x^{-1/2}\). At \(x=4\), this is \(3/(4\cdot2)=3/8\), as obtained above.

Worked Example: A Trigonometric Parametrization

Let \(x(t)=\sin t\) and \(y(t)=\cos t\). Evaluate the second derivative at \(t=0\). The condition \(x'(t)=\cos t\ne0\) holds in a neighborhood of zero.

Differentiate the coordinates and form the slope:

x'(t)=\cos t,\qquad y'(t)=-\sin t, \qquad \frac{dy}{dx}=-\tan t.

Differentiate the slope with respect to \(t\), then divide by \(x'(t)\):

\frac{d}{dt}(-\tan t)=-\sec^2 t, \qquad \frac{d^2y}{dx^2} = \frac{-\sec^2 t}{\cos t} = -\sec^3 t.

At \(t=0\), \(\sec 0=1\), so the result is \(-1\). For an independent check, near \(t=0\) the curve is the graph \(y=\sqrt{1-x^2}\). Its first derivative is \(-x(1-x^2)^{-1/2}\), and its second derivative is \(-(1-x^2)^{-3/2}\). At \(x=0\), this equals \(-1\), confirming the result.

A Common Mistake: Stopping Before Changing Variables

A frequent error is to differentiate the slope with respect to \(t\) and report that result as \(d^2y/dx^2\). That gives the rate at which the slope changes as the parameter changes, not as \(x\) changes. These rates are related, but are generally not equal.

For example, in the first worked example the slope is \(2t/3\). The incorrect calculation would be

\frac{d}{dt}\left(\frac{2t}{3}\right)=\frac{2}{3} \quad\Longrightarrow\quad \text{incorrectly, }\frac{d^2y}{dx^2}=\frac{2}{3}.

The correction is to use \(dt/dx=1/(dx/dt)=1/3\):

\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{2t}{3}\right)\frac{dt}{dx} = \frac{2}{3}\cdot\frac{1}{3} = \frac{2}{9}.

A related error is to apply the formula where \(x'=0\). There the division by \(dx/dt\) is undefined, and \(x\) may not be a valid local coordinate. The formula does not license division by zero; a different local analysis is needed at such a parameter value.

Why the Denominator Matters

The factor \(1/(dx/dt)\) records how quickly \(x\) changes as \(t\) changes. If \(x\) changes rapidly with \(t\), a fixed change in the slope per unit of \(t\) corresponds to a smaller change per unit of \(x\). If \(dx/dt=0\), this conversion cannot be made by the formula.

Thus, keep the two derivative operations distinct:

Expression	What is being differentiated?	Role
\(d(dy/dx)/dt\)	The slope with respect to the parameter	Intermediate rate
\(d(dy/dx)/dx\)	The slope with respect to the horizontal coordinate	Second derivative
\(\bigl(d(dy/dx)/dt\bigr)/(dx/dt)\)	The parameter rate converted to an \(x\)-rate	Chain-rule formula

The derivation is useful because it explains both the calculation and its restriction. Whenever a second derivative is expressed parametrically, the \(t\)-derivative of the slope must be converted back to an \(x\)-derivative, and that conversion requires a nonzero \(dx/dt\).

Check Your Understanding

Answer each question by identifying which variable is the independent variable and where the nonzero-denominator condition enters.

State the chain-rule formula for \(d^2y/dx^2\) in terms of the parameter \(t\).
Why does \(dx/dt\ne0\) allow \(x\) to serve locally as the independent variable?
For \(x(t)=2t\) and \(y(t)=t^2\), find \(dy/dx\) and then \(d^2y/dx^2\) using the derived formula.
Explain why \(d(dy/dx)/dt\) alone is generally not the second derivative with respect to \(x\).
Does regularity of the derivative pair \((x'(t),y'(t))\) by itself guarantee that the formula applies? Explain.