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Computing Partial Derivatives · Tutorial 316 of 999

Third-Order Partial Derivatives

Practice computing pure and mixed third-order partial derivatives while keeping each differentiation step explicit.

Beginner 9 min read

What You'll Learn

  • Interpret a third-order partial as three successive differentiations.
  • Compute pure third partials such as \(f_{yyy}\).
  • Compute mixed third partials such as \(f_{xxy}\).
  • Apply power, product, and quotient rules at each stage.
  • Keep track of which variable is differentiated and which is held fixed.
  • Avoid rearranging the differentiation order without a theorem that permits it.

From Second Partials to Third Partials

A third-order partial derivative is obtained by differentiating a function three times, with respect to one or more variables. Each differentiation acts on the function produced by the previous step. This makes the calculation a sequence of ordinary partial differentiations: take one partial, then take a partial of that result, then take a partial once more.

For example, to calculate \(f_{xxy}\), first differentiate \(f\) with respect to \(x\), differentiate the result with respect to \(x\) again, and then differentiate that result with respect to \(y\). At every step, the other variable is held fixed.

Definition: A third-order partial derivative is a partial derivative formed by applying three first-partial differentiations successively. In \(f_{xxy}\), the indicated sequence is \(x\), then \(x\), then \(y\). In \(f_{yyy}\), all three differentiations are with respect to \(y\). The derivative must exist at each stage for the resulting third partial to be defined.

One way to make the sequence visible is to write each intermediate derivative separately. For instance, \(f_x\) is the first result, \(f_{xx}\) is obtained by differentiating \(f_x\) with respect to \(x\), and \(f_{xxy}\) is obtained by differentiating \(f_{xx}\) with respect to \(y\). The subscripts record the successive variables; they are not instructions to skip the intermediate steps.

Why Repeated Differentiation Works

For a function of two variables, let \(D_x g\) denote the partial derivative of a function \(g\) with respect to \(x\), and let \(D_y g\) denote its partial derivative with respect to \(y\). Thus \(D_x\) and \(D_y\) are operations on functions. For example, the calculation of \(f_{xxy}\) is the operation \(D_y(D_x(D_x f))\).

Justification: A first partial derivative is defined by differentiating the current function with respect to one coordinate while holding the other coordinate fixed. After that operation, the result is itself a function of the coordinates. The same definition therefore applies to the result: it can be differentiated again with respect to a specified coordinate. Repeating this reasoning three times gives exactly three successive partial differentiations. The chain of intermediate functions is why the rules for first partial derivatives can be applied at each stage.

This justification establishes how to compute the indicated sequence; it does not say that different sequences give the same answer. For example, \(f_{xxy}\) and \(f_{xyx}\) correspond to different orders of differentiation. Do not change the order during a calculation unless an applicable result justifies doing so.

Worked Example: A Polynomial with Mixed and Pure Third Partials

Worked Example: A Polynomial with Mixed and Pure Third Partials

Let \(f(x,y)=x^3y^2+4xy^3-5y\). Find \(f_{xxy}\) and \(f_{yyy}\), showing the intermediate derivatives.

For \(f_{xxy}\), differentiate first with respect to \(x\). Treat each power of \(y\) as constant during this step:

$$ f_x=3x^2y^2+4y^3. $$

Differentiate this result with respect to \(x\) again:

$$ f_{xx}=6xy^2. $$

Now differentiate \(f_{xx}\) with respect to \(y\):

$$ f_{xxy}=\frac{\partial}{\partial y}(6xy^2)=12xy. $$

For \(f_{yyy}\), start again from \(f\), and differentiate with respect to \(y\) at each stage:

$$ f_y=2x^3y+12xy^2-5, \qquad f_{yy}=2x^3+24xy, \qquad f_{yyy}=24x. $$

To check the final steps, differentiating \(6xy^2\) with respect to \(y\) gives \(12xy\), as claimed. Differentiating \(2x^3+24xy\) with respect to \(y\) gives \(24x\), also as claimed. In both calculations, terms that are constant with respect to the current variable contribute zero.

Mixed Derivatives: Preserve the Sequence

Mixed third partials involve more than one variable. Write out the first and second partials before taking the third derivative. This is especially helpful when the function contains products, exponentials, or trigonometric factors: a factor that depends on the variable being differentiated must be differentiated, while a factor independent of that variable is held constant.

Worked Example: A Mixed Partial with Exponential and Trigonometric Factors

Let \(f(x,y)=e^{2x}\sin y+x^2\cos y\). Find \(f_{xxy}\). The order is \(x\), then \(x\), then \(y\).

First differentiate with respect to \(x\). Since \(y\) is held fixed, \(\sin y\) and \(\cos y\) are constant factors. Apply the chain rule to \(e^{2x}\):

$$ f_x=2e^{2x}\sin y+2x\cos y. $$

Differentiate this expression with respect to \(x\) again:

$$ f_{xx}=4e^{2x}\sin y+2\cos y. $$

Finally, differentiate with respect to \(y\), holding \(x\) fixed. The derivative of \(\sin y\) is \(\cos y\), and the derivative of \(\cos y\) is \(-\sin y\):

$$ f_{xxy}=4e^{2x}\cos y-2\sin y. $$

Check the last step directly from the displayed \(f_{xx}\): differentiating \(4e^{2x}\sin y\) with respect to \(y\) gives \(4e^{2x}\cos y\), and differentiating \(2\cos y\) gives \(-2\sin y\). Their sum matches the answer.

Pure Third Partials and Repeated Product Rules

A pure third partial differentiates with respect to the same variable three times. Each round still acts on the whole expression obtained in the preceding round. A product may require the product rule again, even if the first differentiation looked simple.

Worked Example: A Pure Third Partial of a Product

Let \(f(x,y)=x^2ye^y\), and find \(f_{yyy}\). During these differentiations, \(x^2\) is constant, so it can be carried along while the product \(ye^y\) is differentiated.

Use the product rule for the first derivative with respect to \(y\):

$$ f_y=x^2\left(e^y+ye^y\right)=x^2(1+y)e^y. $$

For the second derivative, apply the product rule to \((1+y)e^y\):

$$ f_{yy}=x^2\left(e^y+(1+y)e^y\right)=x^2(y+2)e^y. $$

Apply the product rule once more:

$$ f_{yyy}=x^2\left(e^y+(y+2)e^y\right)=x^2(y+3)e^y. $$

As a check, differentiating the displayed \(f_{yy}=x^2(y+2)e^y\) gives \(x^2[e^y+(y+2)e^y]=x^2(y+3)e^y\). This agrees with the third partial above.

Keep the Current Variable in View

At each step, identify the variable being differentiated before using a rule. In \(f_{xxy}\), the first two steps use \(x\), and only the last step uses \(y\). That distinction affects which factors change. It also determines whether a term vanishes because it is constant with respect to the current variable.

Practical rule: Write the intermediate partials in order, and label each one. Do not treat a third partial as a single shortcut calculation; compute the next first partial from the expression currently in front of you.

A common mistake is to differentiate a factor that is constant with respect to the variable at that step. For the polynomial in the first example, the first derivative is \(f_x=3x^2y^2+4y^3\). An incorrect calculation might then claim \(f_{xx}=6xy^2+12y^2\), by differentiating \(4y^3\) as though the second derivative were with respect to \(y\). But the second subscript in \(f_{xx}\) is \(x\), so \(y\) is held fixed and \(4y^3\) has derivative zero. The correct result is \(f_{xx}=6xy^2\). The next step, with respect to \(y\), gives \(f_{xxy}=12xy\).

This example also illustrates why the order must be explicit. The correct result above follows the sequence \(x,x,y\). A different sequence calls for differentiating the appropriate intermediate function with respect to a different variable. Do not assume that changing the sequence leaves the answer unchanged; use only the order specified in the notation unless a result that permits rearrangement has been established.

Why Third Partials Are Useful

Third partial derivatives describe how second partial derivatives change as the input moves in a coordinate direction. For instance, \(f_{xxy}\) measures the change with respect to \(y\) of the second partial \(f_{xx}\). This makes third partials useful when studying how curvature-related quantities vary across a function’s domain, and when computing higher-order terms in local approximations.

For computation, the main discipline is modest but important: start from the original function, take the specified first partial, and repeat until three differentiations have been completed. Applying the familiar first-partial rules carefully at each stage is more reliable than trying to compress the work into one step.

Check Your Understanding

For each question, keep the differentiation sequence visible and show the intermediate partials.

  1. For \(g(x,y)=x^2y^3+2xy\), find \(g_{xxy}\).
  2. For \(h(x,y)=x^3+\sin y\), find \(h_{yyy}\).
  3. For \(p(x,y)=e^x y^2\), find \(p_{xyy}\), differentiating in the order indicated by the subscripts.
  4. In calculating \(f_{xxy}\), which variable is held fixed during the second differentiation?
  5. Why should you not change the order of differentiation in a mixed third partial without justification?

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