Tutorials › Calculus III › Deriving Pure Second Partials by Repeated Differentiation

Home  /  Tutorials  /  Computing Partial Derivatives

Computing Partial Derivatives · Tutorial 314 of 999

Deriving Pure Second Partials by Repeated Differentiation

You will learn to find and check second partial derivatives by repeating a partial differentiation in the same variable.

Beginner 9 min read

What You'll Learn

  • Define pure second partial derivatives by differentiating a first partial again in the same variable.
  • Track how powers of the differentiation variable change on the second pass.
  • Keep factors independent of that variable unchanged.
  • Compute second partials of sums and products with separated coordinate factors.
  • Check each result by differentiating the corresponding first partial again.
  • Recognize and correct errors caused by changing a factor that should be held fixed.

From a First Partial to a Second Partial

A first partial derivative measures how a function changes as one coordinate varies while the other coordinates are held fixed. A pure second partial derivative repeats that operation in the same coordinate. For example, first differentiate with respect to \(x\), then differentiate the result with respect to \(x\) again.

This repeated process records how the rate of change in one coordinate varies as that same coordinate changes. It is a calculation with a precise condition: the first partial must itself be differentiable with respect to the chosen variable at the point under consideration.

Definition: If \(f_x\) is differentiable with respect to \(x\), the pure second partial derivative with respect to \(x\) is \(f_{xx}=\frac{\partial}{\partial x}(f_x)\). If \(f_y\) is differentiable with respect to \(y\), the pure second partial derivative with respect to \(y\) is \(f_{yy}=\frac{\partial}{\partial y}(f_y)\). The notation \(f_{xx}\) means differentiate twice with respect to \(x\); \(f_{yy}\) means differentiate twice with respect to \(y\).

More explicitly, at a point \((x,y)\), the definition of \(f_{xx}\) is the one-variable derivative of the function \(f_x\), with \(y\) fixed:

$$ f_{xx}(x,y) = \lim_{h\to 0} \frac{f_x(x+h,y)-f_x(x,y)}{h}, $$

provided this limit exists. Similarly, \(f_{yy}(x,y)\) is obtained from the difference quotient for \(f_y\) as \(y\) changes while \(x\) stays fixed. The subscript is a reminder of the variable used on that differentiation pass.

How Powers Change on the Second Pass

For a nonnegative integer \(n\), differentiating \(x^n\) once gives \(n x^{n-1}\) when \(n\geq 1\). Differentiating this result with respect to \(x\) again gives \(n(n-1)x^{n-2}\) when \(n\geq 2\). When \(n=0\) or \(n=1\), the second derivative is zero.

Rule: For a nonnegative integer \(n\), $$ \frac{\partial^2}{\partial x^2}x^n = \begin{cases} n(n-1)x^{n-2}, & n\geq 2,\\ 0, & n=0\text{ or }n=1. \end{cases} $$ When differentiating with respect to \(y\), the same rule applies to powers of \(y\).

A factor involving only the other coordinate is constant during both differentiations. Thus, if \(c\) is independent of \(x\), then differentiating \(c x^n\) twice with respect to \(x\) leaves \(c\) as a factor. For example, \(y^3\) is independent of \(x\), so it remains unchanged when the differentiation variable is \(x\).

Worked Example: Finding \(f_{xx}\) Term by Term

Worked Example: Finding \(f_{xx}\) Term by Term

Let \(f(x,y)=3x^4y^2-5x^2y+7y^3+4x\). Find \(f_{xx}\).

First differentiate each term with respect to \(x\). The factors \(y^2\), \(y\), and \(y^3\) do not depend on \(x\), so they stay in place:

$$ \begin{aligned} f_x(x,y) &=3(4x^3)y^2-5(2x)y+7(0)+4\\ &=12x^3y^2-10xy+4. \end{aligned} $$

Now differentiate this first partial with respect to \(x\) again. In \(12x^3y^2\), the factor \(y^2\) remains unchanged; in \(-10xy\), the factor \(y\) remains unchanged; and the constant \(4\) differentiates to zero:

$$ \begin{aligned} f_{xx}(x,y) &=12(3x^2)y^2-10(1)y+0\\ &=36x^2y^2-10y. \end{aligned} $$

To check, differentiate the displayed first partial \(f_x=12x^3y^2-10xy+4\) with respect to \(x\): its terms give \(36x^2y^2\), \(-10y\), and \(0\), respectively. This reproduces \(f_{xx}=36x^2y^2-10y\).

Why the Repeated Power Rule Works

The reason a factor in the other coordinate stays unchanged is built into the definition of a partial derivative: only one coordinate varies at a time. Fix a value \(y=y_0\), and consider the one-variable function \(g(x)=f(x,y_0)\). By the definition of the first partial, \(g'(x)=f_x(x,y_0)\) wherever that derivative exists. If \(g'\) is differentiable, then differentiating it once more gives \(g''(x)=f_{xx}(x,y_0)\). The calculation of a pure second partial is therefore an ordinary second derivative of a coordinate slice.

For a monomial \(c x^n y^m\), where \(c\) is constant and \(m,n\) are nonnegative integers, fixing \(y\) makes \(c y^m\) a constant. The first differentiation gives \(c y^m n x^{n-1}\) for \(n\geq 1\); differentiating again gives \(c y^m n(n-1)x^{n-2}\) for \(n\geq 2\). If \(n=0\), the term is constant in \(x\), and if \(n=1\), its first derivative is constant in \(x\). In either of those cases, the second derivative is zero.

Result: For a finite sum of monomials \(c x^n y^m\), compute \(f_{xx}\) by differentiating each term twice with respect to \(x\). A term with \(n\geq 2\) contributes \(c n(n-1)x^{n-2}y^m\); a term with \(n=0\) or \(n=1\) contributes zero. For \(f_{yy}\), use the exponent of \(y\) in the same way.

This rule applies term by term because partial differentiation is linear: the derivative of a finite sum is the sum of the derivatives. For products, a useful special case is a product of factors separated by coordinate: one factor depends only on \(x\), and the other only on \(y\). On an \(x\)-differentiation pass, the \(y\)-factor is constant, and on a \(y\)-differentiation pass, the \(x\)-factor is constant.

Worked Example: Finding \(f_{yy}\)

Worked Example: Finding \(f_{yy}\)

Let \(f(x,y)=2x^3y^5-4x^2y^2+6x+9y^3\). Find \(f_{yy}\).

Differentiate with respect to \(y\). The factors \(x^3\), \(x^2\), and \(x\) are independent of \(y\):

$$ \begin{aligned} f_y(x,y) &=2x^3(5y^4)-4x^2(2y)+0+9(3y^2)\\ &=10x^3y^4-8x^2y+27y^2. \end{aligned} $$

Differentiate that result with respect to \(y\) again:

$$ \begin{aligned} f_{yy}(x,y) &=10x^3(4y^3)-8x^2(1)+27(2y)\\ &=40x^3y^3-8x^2+54y. \end{aligned} $$

Check by differentiating \(f_y=10x^3y^4-8x^2y+27y^2\) with respect to \(y\): the three terms become \(40x^3y^3\), \(-8x^2\), and \(54y\). The result agrees with the computed \(f_{yy}\).

Products with Separate Coordinate Factors

A product does not automatically make both factors variable-dependent for the chosen differentiation. If \(f(x,y)=a(x)b(y)\), where \(a\) depends only on \(x\) and \(b\) only on \(y\), then \(b(y)\) is constant while differentiating with respect to \(x\), and \(a(x)\) is constant while differentiating with respect to \(y\). Repeating the differentiation therefore gives \(f_{xx}=a''(x)b(y)\) and \(f_{yy}=a(x)b''(y)\), wherever the corresponding one-variable second derivatives exist.

This conclusion depends on the stated separation of the factors. If both factors depend on the differentiation variable, neither can simply be treated as constant. In that situation, the product rule is needed on each differentiation pass. The next example uses separated factors so the role of a coordinate-independent factor is visible.

Worked Example: A Product of Separate Factors

Worked Example: A Product of Separate Factors

Let \(f(x,y)=(x^3-2x)(y^2+1)\). Find both pure second partial derivatives.

For \(f_{xx}\), the factor \(y^2+1\) is independent of \(x\). The first partial is

$$ \begin{aligned} f_x(x,y) &=(3x^2-2)(y^2+1). \end{aligned} $$

Differentiate this with respect to \(x\) again, leaving \(y^2+1\) unchanged:

$$ \begin{aligned} f_{xx}(x,y) &=(6x)(y^2+1)\\ &=6x(y^2+1). \end{aligned} $$

For \(f_{yy}\), the factor \(x^3-2x\) is independent of \(y\). First differentiate once:

$$ \begin{aligned} f_y(x,y) &=(x^3-2x)(2y). \end{aligned} $$

Differentiate again with respect to \(y\):

$$ \begin{aligned} f_{yy}(x,y) &=(x^3-2x)(2)\\ &=2(x^3-2x). \end{aligned} $$

Check both results from the first partials. Differentiating \(f_x=(3x^2-2)(y^2+1)\) with respect to \(x\) gives \(6x(y^2+1)\). Differentiating \(f_y=2y(x^3-2x)\) with respect to \(y\) gives \(2(x^3-2x)\). Both checks match.

A Common Mistake: Differentiating the Fixed Coordinate

Consider \(f(x,y)=x^2y^3\). A common error is to find \(f_x=2xy^3\), then incorrectly differentiate \(y^3\) while taking the next \(x\)-partial:

$$ \text{Incorrect:}\qquad \frac{\partial}{\partial x}(2xy^3) =2x(3y^2) =6xy^2. $$

This calculation differentiates with respect to \(y\), not \(x\). During an \(x\)-partial, \(y^3\) is constant. The correct second partial is

$$ \begin{aligned} f_x&=2xy^3,\\ f_{xx}&=\frac{\partial}{\partial x}(2xy^3)=2y^3. \end{aligned} $$

To check, differentiate the first partial \(2xy^3\) with respect to \(x\): its coefficient \(2y^3\) is constant, so the derivative is exactly \(2y^3\). Keeping the differentiation variable visible at each step prevents this mistake.

A Reliable Calculation Process

For either pure second partial, write the variable for each differentiation pass before doing the algebra. Then apply the first partial rules to obtain an intermediate expression, and differentiate that expression once more in the same variable. A short check against the intermediate first partial is useful because the second differentiation must act on that expression, not on the original function in a different variable.

1
Choose the variable.
For \(f_{xx}\), use \(x\) on both passes; for \(f_{yy}\), use \(y\) on both passes.
2
Find the first partial.
Differentiate \(f\) with respect to the chosen variable, holding the other coordinate fixed.
3
Differentiate the result again.
Apply the same variable and the relevant rules to the first partial.
4
Check the calculation.
Differentiate the written first partial once more and compare term by term with the proposed second partial.

Why Pure Second Partials Matter

Pure second partials describe how a coordinatewise rate of change itself changes along that coordinate. They are important in studying the local shape of functions of several variables and in later calculations involving second-order information. The computation here is deliberately coordinate-specific: \(f_{xx}\) repeats the \(x\)-partial, while \(f_{yy}\) repeats the \(y\)-partial. Differentiating first in one coordinate and then in another is a different operation.

The essential habits are straightforward: identify the variable, hold the other coordinate fixed, differentiate the first partial again, and verify the result from that first partial. These habits make repeated differentiation both systematic and easier to audit.

Check Your Understanding

For each question, state which coordinate is held fixed during the differentiations.

  1. For \(f(x,y)=5x^3y^2\), find \(f_x\) and then \(f_{xx}\).
  2. For \(g(x,y)=4x^2y^4-3y\), find \(g_y\) and then \(g_{yy}\).
  3. If \(h(x,y)=a(x)b(y)\), with the indicated factors depending only on their displayed variables, what are \(h_{xx}\) and \(h_{yy}\) in terms of \(a\) and \(b\)?
  4. Explain why differentiating \(2xy^3\) with respect to \(x\) gives \(2y^3\), rather than \(6xy^2\).
  5. What condition must hold for \(f_{xx}(x,y)\) to be defined by repeated differentiation at a point?

Previous

Next

Back to Tutorials