Clairaut's Theorem - Multivariable Calculus

1. Let $f(x,y)$ be a function with continuous second-order partial derivatives on $\mathbb{R}^2$. Suppose $f_{xx}(x,y) = 6x y^2$, $f_{yy}(x,y) = e^y \cos(x)$, and $f_{xy}(x,y) = x^3 + 2y \sin(x)$. What is the value of $f_{yx}(1, 0)$?
   Answer: 1 Clairaut's theorem states that if a function's second-order partial derivatives are continuous on a disk containing a point, then the mixed partial derivatives are equal at that point. Since $f$ has continuous second-order partial derivatives, we have $f_{yx}(x,y) = f_{xy}(x,y)$ for all $(x,y)$. Therefore, we only need to evaluate the given expression for $f_{xy}$ at the point $(1,0)$. The information about $f_{xx}$ and $f_{yy}$ is extraneous. $f_{yx}(1,0) = f_{xy}(1,0) = (1)^3 + 2(0)\sin(1) = 1 + 0 = 1$.
2. Let $z(x,y) = f(x^2 + y^2)$ where $f$ is a twice continuously differentiable function of one variable. Using Clairaut's theorem, which equation must $f$ satisfy for the mixed partials of $z$ to be equal?
   Answer: No additional condition on $f$ is needed; Clairaut's theorem automatically applies since $z$ inherits twice continuous differentiability from $f$ Since $f$ is twice continuously differentiable and $z(x,y) = f(x^2 + y^2)$ is a composition of $f$ with the smooth function $x^2 + y^2$, the function $z$ automatically inherits twice continuous differentiability everywhere. By the chain rule, $z_{xy} = 2xyf''(x^2 + y^2)$ and $z_{yx} = 2xyf''(x^2 + y^2)$, so they are automatically equal. Clairaut's theorem applies without additional conditions on $f$. The other choices incorrectly suggest additional constraints are needed when the composition already ensures the required regularity and symmetry.
3. A function $f(x, y)$ has continuous second partial derivatives, and it is known that $f_{xy}(x, y) = 6x + 4y$. If $g(x, y) = f(y, x)$, what is $g_{xy}(x, y)$?
   Answer: $6y + 4x$ Let $u=y$ and $v=x$. Then $g(x,y) = f(u,v)$. Using the chain rule, we find the partial derivatives of $g$. $g_x(x,y) = f_u(u,v) \frac{\partial u}{\partial x} + f_v(u,v) \frac{\partial v}{\partial x} = f_u(y,x) \cdot 0 + f_v(y,x) \cdot 1 = f_v(y,x)$. Now we differentiate $g_x$ with respect to $y$ to find $g_{xy}$. $g_{xy}(x,y) = \frac{\partial}{\partial y}(f_v(y,x))$. Applying the chain rule again: $g_{xy}(x,y) = f_{vu}(y,x) \frac{\partial u}{\partial y} + f_{vv}(y,x) \frac{\partial v}{\partial y} = f_{vu}(y,x) \cdot 1 + f_{vv}(y,x) \cdot 0 = f_{vu}(y,x)$. Since $f$ has continuous second partial derivatives, Clairaut's theorem states that $f_{vu} = f_{uv}$. The variables for $f$ are $u$ and $v$, so this is $f_{xy}$ in its original variable names. Therefore, $f_{vu}(u,v) = f_{uv}(u,v) = 6u + 4v$. We need to evaluate this at $(u,v) = (y,x)$. $g_{xy}(x,y) = f_{vu}(y,x) = 6y + 4x$.
4. Consider $g(x,y) = e^{x^2-y^2} \cos(xy)$. If you need to compute both $g_{xy}$ and $g_{yx}$ to verify they satisfy Clairaut's theorem, which approach is most efficient?
   Answer: First verify that $g$ is twice continuously differentiable, then compute only $g_{xy}$ and apply Clairaut's theorem for $g_{yx}$ Since $g(x,y) = e^{x^2-y^2} \cos(xy)$ is composed of exponential and trigonometric functions with polynomial arguments, it is infinitely differentiable everywhere. Once we verify this (which is immediate for such functions), Clairaut's theorem guarantees $g_{xy} = g_{yx}$, so we only need to compute one of them. Choice A assumes the theorem applies without verification. Choice B is inefficient since it computes both when one suffices. Choice D incorrectly invokes 'symmetry properties' that don't actually apply to this function form.
5. A function $f(x,y)$ has continuous third-order partial derivatives everywhere. It is known that $f_{xxy}(x,y) = 12xy + e^x$ and $f_{yxx}(x,y) = 12xy + e^x$. What is $f_{xyx}(x,y)$?
   Answer: $12xy + e^x$ Clairaut's theorem extends to higher-order derivatives. If a function's third-order partial derivatives are all continuous, then the order of differentiation does not matter for any third-order mixed partial derivative. This means that any partial derivative with two 'x' differentiations and one 'y' differentiation will be the same, regardless of the order. Therefore, $f_{xxy} = f_{yxx} = f_{xyx}$. Since we are given that $f_{xxy}(x,y) = 12xy + e^x$, it must be that $f_{xyx}(x,y)$ is also equal to $12xy + e^x$. The information about $f_{yxx}$ is redundant and serves to confirm this fact.
6. Suppose there exists a function $f(x,y)$ with continuous second-order partial derivatives such that $f_x(x,y) = 2xy^k + y^2$ and $f_y(x,y) = 3x^2y^2 + 2xy$. What must be the value of the constant $k$?
   Answer: 3 If a function $f(x,y)$ exists with continuous second-order partial derivatives, then by Clairaut's theorem, we must have $(f_x)_y = (f_y)_x$. We compute these two mixed partial derivatives. $(f_x)_y = \frac{\partial}{\partial y}(2xy^k + y^2) = 2kxy^{k-1} + 2y$. $(f_y)_x = \frac{\partial}{\partial x}(3x^2y^2 + 2xy) = 6xy^2 + 2y$. For these expressions to be equal for all $(x,y)$, the coefficients of like terms must match. $2kxy^{k-1} + 2y = 6xy^2 + 2y$. This implies $2kxy^{k-1} = 6xy^2$. Comparing the powers of $y$, we see that $k-1 = 2$, which gives $k=3$. Comparing the coefficients, $2k = 6$, which also gives $k=3$. Thus, the only possible value is $k=3$.
7. For the function $u(x,y) = \ln(x^2 + y^2 + 1) + \arctan\left(\frac{y}{x}\right)$, Clairaut's theorem can be applied to conclude $u_{xy} = u_{yx}$ everywhere except possibly where:
   Answer: $x = 0$, where the arctangent function creates discontinuities that violate the continuity requirements of Clairaut's theorem The function $\ln(x^2 + y^2 + 1)$ is well-defined and smooth everywhere since $x^2 + y^2 + 1 \geq 1 > 0$. However, $\arctan(y/x)$ is undefined when $x = 0$, and the partial derivatives involving this term will have discontinuities along the $y$-axis. This violates the continuity condition needed for Clairaut's theorem. Choice A is impossible since $x^2 + y^2 + 1 \geq 1$. Choice C incorrectly identifies where problems occur. Choice D incorrectly suggests both terms create issues and misidentifies the problematic region.
8. Let $f(x,y)$ be a $C^2$ function defined on $\mathbb{R}^2$. If the mixed partial derivative $f_{xy}$ is given by the expression $f_{xy}(x,y) = e^{x^2} \cos(y^2)$, which of the following expressions represents $f_{yx}(x,y)$?
   Answer: $e^{x^2} \cos(y^2)$ When you encounter mixed partial derivatives in multivariable calculus, the key concept to remember is Clairaut's theorem (also called Schwarz's theorem). This fundamental result states that for any $C^2$ function, the mixed partial derivatives are equal: $f_{xy} = f_{yx}$ Since we're told that $f(x,y)$ is $C^2$ (meaning it has continuous second partial derivatives) and given that $f_{xy}(x,y) = e^{x^2} \cos(y^2)$, Clairaut's theorem immediately tells us that $f_{yx}(x,y)$ must equal the same expression. Therefore, $f_{yx}(x,y) = e^{x^2} \cos(y^2)$, which is answer choice C. Let's examine why the other options are incorrect. Choice A ($-2y e^{x^2} \sin(y^2)$) appears to be the result of taking a partial derivative of $f_{xy}$ with respect to $y$, giving you a third-order derivative rather than the mixed second derivative. Choice B ($2x e^{x^2} \cos(y^2)$) similarly looks like the partial derivative of $f_{xy}$ with respect to $x$. Choice D (0) would only be correct if the mixed partials were somehow always zero, which has no basis in the given information. Study tip: Whenever you see a problem involving mixed partial derivatives of a $C^2$ function, immediately think "equality of mixed partials." This theorem is one of the most frequently tested concepts in multivariable calculus, so memorize it: for smooth functions, $f_{xy} = f_{yx}$ always.
9. A function $h(x,y)$ satisfies $h_{xy}(2,3) = 7$ and both $h_{xy}$ and $h_{yx}$ are continuous in a neighborhood of $(2,3)$. If $h_{yx}$ is computed directly and found to equal $5$ at $(2,3)$, what conclusion should be drawn?
   Answer: There must be a computational error since Clairaut's theorem guarantees $h_{xy}(2,3) = h_{yx}(2,3) = 7$ under the given conditions If both mixed partial derivatives exist and are continuous in a neighborhood of $(2,3)$, then by Clairaut's theorem they must be equal at that point. Since we're told both conditions are satisfied, getting different values indicates a computational error. The theorem's conditions are met, so equality must hold. Choice B incorrectly requires global differentiability. Choice C misstates the theorem's requirements. Choice D incorrectly suggests we should doubt the given continuity condition rather than suspect computational error.
10. For $w(x,y) = |xy|$, what prevents the direct application of Clairaut's theorem to conclude $w_{xy} = w_{yx}$ throughout the domain?
    Answer: While $w_{xy}$ and $w_{yx}$ exist almost everywhere, they are not continuous along the lines $x = 0$ and $y = 0$ When analyzing whether Clairaut's theorem applies to a function involving absolute values, you need to check if the mixed partial derivatives exist and are continuous throughout the domain. Clairaut's theorem states that $w_{xy} = w_{yx}$ when both mixed partials are continuous. For $w(x,y) = |xy|$, let's trace what happens. In each quadrant where $xy$ has a consistent sign, the function behaves like $w = xy$ or $w = -xy$, giving us $w_{xy} = w_{yx} = 1$ or $w_{xy} = w_{yx} = -1$ respectively. The mixed partials do exist almost everywhere. The critical issue occurs along the coordinate axes ($x = 0$ or $y = 0$). Here, the mixed partial derivatives exist but experience jump discontinuities as you cross from one quadrant to another. Since Clairaut's theorem requires the mixed partials to be continuous, this discontinuity prevents direct application of the theorem. Option A is incorrect because the mixed partials do exist along the coordinate axes through limit processes. Option B misses the point—the mixed partials are equal within each quadrant where they're continuous. Option C is wrong because $|xy|$ is continuous everywhere, including at the origin ($\lim_{(x,y) \to (0,0)} |xy| = 0 = w(0,0)$). Option D correctly identifies that while the mixed partials exist almost everywhere, their lack of continuity along the coordinate axes prevents applying Clairaut's theorem. Study tip: For absolute value functions in multivariable calculus, always check continuity of derivatives at boundary lines where the expression inside the absolute value changes sign.
11. Let $f(x,y) = x^3y^2 + e^{xy} \sin(x^2y)$. If $f_{xy}(a,b)$ exists and is continuous in a neighborhood of $(a,b)$, which of the following statements about $f_{yx}(a,b)$ is necessarily true?
    Answer: $f_{yx}(a,b) = f_{xy}(a,b)$ provided that $f_{yx}(a,b)$ exists and is continuous in a neighborhood of $(a,b)$ Clairaut's theorem states that if both mixed partial derivatives exist and are continuous in a neighborhood of a point, then they are equal at that point. The correct answer requires both conditions: existence and continuity of both mixed partials. Choice A incorrectly assumes elementary functions automatically satisfy all conditions. Choice C incorrectly suggests the function type matters more than continuity. Choice D ignores the continuity requirement entirely.
12. A function $g(x,y)$ has continuous second-order partial derivatives ($g$ is $C^2$). At the point $P=(3,4)$, the gradient is $\nabla g(P) = \langle 1, -2 \rangle$ and its Hessian matrix is $H_g(P) = \begin{pmatrix} 5 & A \\ 8 & -1 \end{pmatrix}$. What must be the value of $A$?
    Answer: 8 The Hessian matrix of a function $g(x,y)$ is defined as $H_g = \begin{pmatrix} g_{xx} & g_{xy} \\ g_{yx} & g_{yy} \end{pmatrix}$. The problem states that $g$ is a $C^2$ function, which means its second-order partial derivatives are continuous. By Clairaut's theorem, this implies that the mixed partial derivatives are equal: $g_{xy} = g_{yx}$. Therefore, the Hessian matrix must be symmetric. From the given matrix, we have $g_{xy}(P) = A$ and $g_{yx}(P) = 8$. For the matrix to be symmetric, we must have $A=8$. The information about the gradient is extraneous.
13. A student computes $f_{xy}$ for $f(x,y) = x^3y^2 + 2xy$ and gets $6x^2y + 2$. Then computes $f_{yx}$ and gets $6xy^2 + 2$. The student concludes that Clairaut's theorem fails for this function. What is the most likely explanation?
    Answer: The student made an error in computing $f_{yx}$; the correct value should be $6x^2y + 2$ to match $f_{xy}$ For $f(x,y) = x^3y^2 + 2xy$: $f_x = 3x^2y^2 + 2y$, so $f_{xy} = 6x^2y + 2$. Also, $f_y = 2x^3y + 2x$, so $f_{yx} = 6x^2y + 2$. The student correctly computed $f_{xy}$ but incorrectly computed $f_{yx}$. Since this is a polynomial, both mixed partials exist and are continuous everywhere, so Clairaut's theorem guarantees they are equal. Choice A incorrectly suggests the theorem doesn't apply to polynomials. Choice C incorrectly identifies which computation is wrong. Choice D gives an incorrect value for both partials.
14. Let $u(x, t)$ be a $C^2$ function representing the temperature at position $x$ and time $t$, which satisfies the heat equation $u_t = k u_{xx}$ for some constant $k$. Which of the following quantities must be equal to $u_{tx}$?
    Answer: $u_{xt}$ The problem states that $u(x,t)$ is a $C^2$ function, which means its second-order partial derivatives are continuous. Clairaut's theorem states that for such functions, the order of mixed partial differentiation does not matter. Therefore, $u_{tx} = \frac{\partial}{\partial x}(\frac{\partial u}{\partial t})$ must be equal to $u_{xt} = \frac{\partial}{\partial t}(\frac{\partial u}{\partial x})$. The heat equation itself provides a relationship between different partials, but the equality of mixed partials comes directly from the smoothness condition ($C^2$) and Clairaut's theorem.
15. Let $f(x,y)$ be a function with continuous second-order partial derivatives on its domain. Let a change of variables to polar coordinates be given by $x = r\cos\theta$ and $y=r\sin\theta$, and define $g(r, \theta) = f(r\cos\theta, r\sin\theta)$. Which of the following statements must be true for all $r>0$?
    Answer: $g_{r\theta} = g_{\theta r}$ When you encounter problems involving change of variables and mixed partial derivatives, the key concept to recall is Clairaut's theorem (also called Schwarz's theorem), which states that mixed partial derivatives are equal when the function has continuous second-order partial derivatives. Since $f(x,y)$ has continuous second-order partial derivatives and $g(r,\theta) = f(r\cos\theta, r\sin\theta)$, the composite function $g$ inherits this smoothness property. The transformation from Cartesian to polar coordinates involves smooth functions ($\cos\theta$ and $\sin\theta$), so the chain rule applies cleanly, and all second-order partial derivatives of $g$ remain continuous. By Clairaut's theorem, this continuity guarantees that $g_{r\theta} = g_{\theta r}$ for all points where both mixed partials exist and are continuous, which includes all $r > 0$. Looking at the wrong answers: Choice A incorrectly suggests the relationship depends on the specific function $f$, but Clairaut's theorem applies universally to any function with continuous second partials. Choice B claims $g_{r\theta} = -g_{\theta r}$, which would violate Clairaut's theorem and has no mathematical basis. Choice C states $g_{rr} + g_{\theta\theta} = 0$, which resembles a Laplace equation but isn't generally true for arbitrary functions under polar coordinate transformation. Study tip: Whenever you see mixed partial derivatives with continuous second-order partials, immediately think "Clairaut's theorem" - the order of differentiation doesn't matter. Coordinate transformations preserve this property as long as the transformation functions are smooth.
16. For a vector field $\mathbf{F}(x,y) = \langle P(x,y), Q(x,y) \rangle$ to be conservative on a simply connected domain, there must exist a potential function $f(x,y)$ such that $\nabla f = \mathbf{F}$. If $P$ and $Q$ have continuous first partial derivatives, this is equivalent to the condition $P_y = Q_x$. This condition is a direct consequence of which theorem?
    Answer: Clairaut's Theorem If a potential function $f$ exists such that $\nabla f = \langle f_x, f_y \rangle = \langle P, Q \rangle$, then $P = f_x$ and $Q = f_y$. The condition that $P$ and $Q$ have continuous first partial derivatives implies that $f$ has continuous second partial derivatives. Clairaut's theorem on the equality of mixed partials states that under these conditions, $f_{xy} = f_{yx}$. Differentiating $P$ with respect to $y$ gives $P_y = (f_x)_y = f_{xy}$. Differentiating $Q$ with respect to $x$ gives $Q_x = (f_y)_x = f_{yx}$. Therefore, the condition $P_y = Q_x$ is a direct consequence of $f_{xy} = f_{yx}$, which is Clairaut's theorem.
17. A function $\phi(x,y)$ is known to have continuous second partial derivatives everywhere except possibly at $(0,0)$. If $\phi_{xy}(0,0)$ and $\phi_{yx}(0,0)$ both exist, what additional condition is needed to conclude $\phi_{xy}(0,0) = \phi_{yx}(0,0)$?
    Answer: That $\phi_{xy}$ and $\phi_{yx}$ are continuous at $(0,0)$, completing the hypotheses required by Clairaut's theorem When you encounter questions about mixed partial derivatives and their equality, you're dealing with Clairaut's theorem (also called Schwarz's theorem), which is fundamental to understanding when the order of partial differentiation doesn't matter. Clairaut's theorem states that if the mixed partial derivatives $\phi_{xy}$ and $\phi_{yx}$ both exist in a neighborhood of a point and are continuous at that point, then they are equal at that point. The key insight is that mere existence of the mixed partials isn't sufficient—continuity at the point is the crucial additional requirement. Option D correctly identifies this missing piece. You're told the mixed partials exist at $(0,0)$, but Clairaut's theorem requires their continuity at $(0,0)$ to guarantee equality. Option A is incorrect because continuity of $\phi$ itself doesn't ensure the mixed partials are equal. A function can be continuous while having discontinuous derivatives. Option B reflects a common misconception. The symmetry of partial differentiation isn't automatic—it requires the conditions of Clairaut's theorem. There are counterexamples where mixed partials exist but aren't equal when continuity fails. Option C is wrong because knowing that $\phi_{xx}$ and $\phi_{yy}$ exist doesn't help establish equality of the mixed partials. The "regularity" that matters specifically concerns the continuity of $\phi_{xy}$ and $\phi_{yx}$. Remember: When you see mixed partial derivatives, immediately think Clairaut's theorem. Existence alone isn't enough—you need continuity of the mixed partials at the point to guarantee their equality.
18. Let $F(x,y) = x^4 + y^4 + x^2y^2 + xy$. To verify that $F_{xy} = F_{yx}$ using Clairaut's theorem, what is the minimum condition that must be verified?
    Answer: That both $F_{xy}$ and $F_{yx}$ exist and are continuous, which for polynomials is automatically satisfied everywhere Clairaut's theorem requires that both mixed partial derivatives exist and are continuous in a neighborhood of the point in question. For polynomial functions, all partial derivatives of any order exist and are continuous everywhere, so this condition is automatically satisfied. Choice A only mentions continuity of $F$ itself, which is insufficient. Choice B incorrectly suggests existence alone is sufficient. Choice D is overly broad, mentioning all second partials when only the mixed partials' continuity is needed for Clairaut's theorem.

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