Partial Derivatives - Multivariable Calculus

1. Consider the function $g(x,y) = \ln(x^2 + y^2) + \arctan(y/x)$ for $x > 0$. What is $\frac{\partial^2 g}{\partial x^2} + \frac{\partial^2 g}{\partial y^2}$ at the point $(1,1)$?
   Answer: $0$ For $\ln(x^2 + y^2)$: $\frac{\partial}{\partial x} = \frac{2x}{x^2+y^2}$, so $\frac{\partial^2}{\partial x^2} = \frac{2(x^2+y^2) - 2x(2x)}{(x^2+y^2)^2} = \frac{2y^2-2x^2}{(x^2+y^2)^2}$. Similarly, $\frac{\partial^2}{\partial y^2} = \frac{2x^2-2y^2}{(x^2+y^2)^2}$. For $\arctan(y/x)$: $\frac{\partial}{\partial x} = \frac{1}{1+(y/x)^2} \cdot \frac{-y}{x^2} = \frac{-y}{x^2+y^2}$, so $\frac{\partial^2}{\partial x^2} = \frac{2xy}{(x^2+y^2)^2}$. And $\frac{\partial^2}{\partial y^2} = \frac{-2xy}{(x^2+y^2)^2}$. Adding all second partials: $\frac{2y^2-2x^2}{(x^2+y^2)^2} + \frac{2x^2-2y^2}{(x^2+y^2)^2} + \frac{2xy}{(x^2+y^2)^2} + \frac{-2xy}{(x^2+y^2)^2} = 0$.
2. Let $f(x, y) = x \cos(y) + y e^{xy}$. What is the value of the third-order partial derivative $f_{yxy}(0, \pi)$?
   Answer: 3 To find $f_{yxy}$, we compute the partial derivatives in order. First with respect to $y$, then $x$, then $y$ again. 1. Differentiate with respect to $y$: $f_y = \frac{\partial}{\partial y}(x \cos(y) + y e^{xy}) = -x \sin(y) + (1 \cdot e^{xy} + y \cdot x e^{xy}) = -x \sin(y) + e^{xy}(1+xy)$. 2. Differentiate $f_y$ with respect to $x$: $f_{yx} = \frac{\partial}{\partial x}(-x \sin(y) + e^{xy} + xye^{xy}) = -\sin(y) + y e^{xy} + (1 \cdot ye^{xy} + x \cdot y^2 e^{xy}) = -\sin(y) + 2ye^{xy} + xy^2e^{xy} = -\sin(y) + e^{xy}(2y+xy^2)$. 3. Differentiate $f_{yx}$ with respect to $y$: $f_{yxy} = \frac{\partial}{\partial y}(-\sin(y) + e^{xy}(2y+xy^2))$. Using the product rule on the second term: $f_{yxy} = -\cos(y) + [(\frac{\partial}{\partial y}e^{xy})(2y+xy^2) + e^{xy}(\frac{\partial}{\partial y}(2y+xy^2))]$ $f_{yxy} = -\cos(y) + [x e^{xy}(2y+xy^2) + e^{xy}(2+2xy)]$. 4. Evaluate at the point $(0, \pi)$: $f_{yxy}(0, \pi) = -\cos(\pi) + [0 \cdot e^0(2\pi+0) + e^0(2+0)] = -(-1) + 0 + 1(2) = 1 + 2 = 3$. Distractor A is incorrect; a value of 1 can result from forgetting the second term of the product rule in the final differentiation step. Distractor B is incorrect; a value of 2 can result from incorrectly differentiating $y e^{xy}$ as just $e^{xy}$ in the first step. Distractor D is incorrect; a value of 0 can result from evaluating a derivative at the point too early in the process and then differentiating the resulting constant.
3. The equation $x^2 z^3 + y \sin(\pi z) = -3$ defines $z$ implicitly as a function of $x$ and $y$. What is the value of $\frac{\partial z}{\partial x}$ at the point $(2, -1, -1)$?
   Answer: $\frac{4}{12 + \pi}$ We use the formula for implicit differentiation for a function $F(x, y, z) = 0$, which is $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$. First, define $F(x, y, z) = x^2 z^3 + y \sin(\pi z) + 3 = 0$. Next, compute the partial derivatives $F_x$ and $F_z$: $F_x = \frac{\partial}{\partial x}(x^2 z^3 + y \sin(\pi z) + 3) = 2xz^3$. $F_z = \frac{\partial}{\partial z}(x^2 z^3 + y \sin(\pi z) + 3) = 3x^2z^2 + y \cos(\pi z) \cdot \pi = 3x^2z^2 + \pi y \cos(\pi z)$. Now, evaluate these partial derivatives at the point $(x, y, z) = (2, -1, -1)$: $F_x(2, -1, -1) = 2(2)(-1)^3 = -4$. $F_z(2, -1, -1) = 3(2^2)(-1)^2 + \pi(-1)\cos(-\pi) = 3(4)(1) - \pi(-1) = 12 + \pi$. Finally, substitute these values into the formula: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{-4}{12 + \pi} = \frac{4}{12 + \pi}$. Distractor A is incorrect; it results from forgetting the chain rule factor of $\pi$ when differentiating $\sin(\pi z)$. Distractor B is incorrect; it results from a sign error in the implicit differentiation formula. Distractor C is incorrect; it results from inverting the formula to $-F_z/F_x$.
4. Let $f(x, y) = \frac{x^3 + y^3}{x-y}$ for $x \neq y$. Which of the following partial differential equations does the function $f$ satisfy?
   Answer: $x f_x + y f_y = 2f$ The function $f(x, y)$ is a homogeneous function. A function is homogeneous of degree $k$ if $f(tx, ty) = t^k f(x, y)$ for any $t > 0$. For the given function: $f(tx, ty) = \frac{(tx)^3 + (ty)^3}{tx-ty} = \frac{t^3(x^3 + y^3)}{t(x-y)} = t^2 \frac{x^3 + y^3}{x-y} = t^2 f(x, y)$. Thus, $f$ is a homogeneous function of degree $k=2$. Euler's Homogeneous Function Theorem states that if $f$ is a differentiable homogeneous function of degree $k$, then it satisfies the partial differential equation $x f_x + y f_y = kf$. For this function, with $k=2$, the equation is $x f_x + y f_y = 2f$. This provides a direct solution without computing the derivatives. Direct computation of $f_x$ and $f_y$ using the quotient rule would be lengthy but would yield the same result. Distractor A is incorrect; $k=3$ corresponds to the degree of the numerator, not the overall function. Distractor C and D are incorrect alternative differential equations that are not satisfied by this function.
5. Let $g(x,y) = f(x^2-y^2, 2xy)$ where $f$ is a differentiable function of two variables. If $f_u(1,0) = 3$ and $f_v(1,0) = -2$, what is $\frac{\partial g}{\partial y}$ at the point $(1,0)$?
   Answer: $-4$ Let $u = x^2 - y^2$ and $v = 2xy$. Using the chain rule: $\frac{\partial g}{\partial y} = f_u \frac{\partial u}{\partial y} + f_v \frac{\partial v}{\partial y}$. We have $\frac{\partial u}{\partial y} = -2y$ and $\frac{\partial v}{\partial y} = 2x$. At the point $(x,y) = (1,0)$, we have $u = 1-0 = 1$ and $v = 0$, so we use $f_u(1,0) = 3$ and $f_v(1,0) = -2$. Also, $\frac{\partial u}{\partial y} = -2(0) = 0$ and $\frac{\partial v}{\partial y} = 2(1) = 2$. Therefore, $\frac{\partial g}{\partial y} = 3(0) + (-2)(2) = -4$.
6. Consider the function $f(x, y)$ defined by $f(x, y) = (x^2 + y^2) \sin\left(\frac{1}{\sqrt{x^2+y^2}}\right)$ for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. What is the value of the partial derivative $f_y(0, 0)$?
   Answer: $0$ To find the partial derivative at the origin for this piecewise function, we must use the limit definition of the partial derivative: $f_y(0, 0) = \lim_{k \to 0} \frac{f(0, 0+k) - f(0, 0)}{k}$. First, find the expression for $f(0, k)$: $f(0, k) = (0^2 + k^2) \sin\left(\frac{1}{\sqrt{0^2+k^2}}\right) = k^2 \sin\left(\frac{1}{|k|}\right)$. Now, substitute this into the limit definition, using $f(0,0)=0$: $f_y(0, 0) = \lim_{k \to 0} \frac{k^2 \sin(1/|k|) - 0}{k} = \lim_{k \to 0} k \sin(1/|k|)$. We can evaluate this limit using the Squeeze Theorem. We know that for any value of $k \neq 0$, $-1 \le \sin(1/|k|) \le 1$. Multiplying the inequality by $k$ (and considering $k>0$ and $k<0$ which results in multiplying by $|k|$): $-|k| \le k \sin(1/|k|) \le |k|$. As $k \to 0$, both $-|k|$ and $|k|$ approach 0. Therefore, by the Squeeze Theorem, the limit must be 0. $f_y(0, 0) = 0$. Distractor D is incorrect. A common mistake is to compute $f_y(x, y)$ using differentiation rules and then try to take the limit as $(x, y) \to (0, 0)$. The resulting expression for $f_y(x, y)$ does not have a limit at the origin, but this does not mean the partial derivative at the origin does not exist. The limit definition must be used.
7. Let $w = x^3 y + y^2 z^2$, where $x = s \cos(t)$, $y = s \sin(t)$, and $z = s t$. What is the value of $\frac{\partial w}{\partial t}$ when $s=2$ and $t=\frac{\pi}{2}$?
   Answer: $16\pi$ We use the chain rule for partial derivatives: $\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial t}$. First, find the point $(x,y,z)$ corresponding to $(s,t) = (2, \pi/2)$: $x = 2 \cos(\pi/2) = 0$ $y = 2 \sin(\pi/2) = 2$ $z = 2(\pi/2) = \pi$ Next, compute the partial derivatives of $w$ with respect to $x, y, z$ and evaluate them at $(0, 2, \pi)$: $\frac{\partial w}{\partial x} = 3x^2 y \implies 3(0)^2(2) = 0$ $\frac{\partial w}{\partial y} = x^3 + 2yz^2 \implies 0^3 + 2(2)(\pi)^2 = 4\pi^2$ $\frac{\partial w}{\partial z} = 2y^2z \implies 2(2)^2(\pi) = 8\pi$ Then, compute the partial derivatives of $x, y, z$ with respect to $t$ and evaluate them at $(s,t) = (2, \pi/2)$: $\frac{\partial x}{\partial t} = -s \sin(t) \implies -2 \sin(\pi/2) = -2$ $\frac{\partial y}{\partial t} = s \cos(t) \implies 2 \cos(\pi/2) = 0$ $\frac{\partial z}{\partial t} = s \implies 2$ Finally, substitute these values into the chain rule formula: $\frac{\partial w}{\partial t} = (0)(-2) + (4\pi^2)(0) + (8\pi)(2) = 0 + 0 + 16\pi = 16\pi$. Distractor A is incorrect; it results from omitting the $\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}$ term from the chain rule. Distractor B is incorrect; it results from computing $\frac{\partial w}{\partial s}$ instead of $\frac{\partial w}{\partial t}$. Distractor D is incorrect; it can result from swapping the derivative formulas for sine and cosine.
8. Let $f(u, v)$ be a function with continuous second partial derivatives. Define $g(x, y) = f(x^2 y, x-y)$. Given that $f_u(0,0)=3$, $f_v(0,0)=2$, $f_{uu}(0,0)=1$, $f_{uv}(0,0)=-1$, and $f_{vv}(0,0)=4$, what is the value of $g_{xy}(0,0)$?
   Answer: $-4$ We need to compute $g_{xy}(0,0) = \frac{\partial}{\partial y}(\frac{\partial g}{\partial x})$ at $(x,y)=(0,0)$. Let $u = x^2 y$ and $v = x-y$. First, find $g_x$ using the chain rule: $g_x = \frac{\partial g}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = f_u(u,v) \cdot (2xy) + f_v(u,v) \cdot (1)$. Next, differentiate $g_x$ with respect to $y$ using the product rule and the chain rule: $g_{xy} = \frac{\partial}{\partial y} (f_u(u,v) \cdot 2xy) + \frac{\partial}{\partial y}(f_v(u,v))$. For the first term: $\frac{\partial}{\partial y} (f_u(u,v) \cdot 2xy) = (\frac{\partial}{\partial y} f_u(u,v)) \cdot (2xy) + f_u(u,v) \cdot (\frac{\partial}{\partial y} 2xy)$. Here, $\frac{\partial}{\partial y} f_u(u,v) = \frac{\partial f_u}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f_u}{\partial v}\frac{\partial v}{\partial y} = f_{uu} (x^2) + f_{uv} (-1)$. So the first term becomes $(f_{uu} x^2 - f_{uv}) \cdot 2xy + f_u \cdot 2x$. For the second term: $\frac{\partial}{\partial y} f_v(u,v) = \frac{\partial f_v}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f_v}{\partial v}\frac{\partial v}{\partial y} = f_{vu} (x^2) + f_{vv} (-1)$. Combining everything: $g_{xy} = (f_{uu} x^2 - f_{uv}) 2xy + 2x f_u + f_{vu} x^2 - f_{vv}$. Now, evaluate at $(x,y)=(0,0)$. At this point, $u = 0^2 \cdot 0 = 0$ and $v = 0-0 = 0$. We use the given derivative values at $(u,v)=(0,0)$. $g_{xy}(0,0) = (f_{uu}(0,0) \cdot 0 - f_{uv}(0,0)) \cdot 0 + 2(0)f_u(0,0) + f_{vu}(0,0) \cdot 0 - f_{vv}(0,0)$. $g_{xy}(0,0) = 0 + 0 + 0 - f_{vv}(0,0) = -f_{vv}(0,0)$. Given $f_{vv}(0,0) = 4$, we have $g_{xy}(0,0) = -4$. Distractor B is incorrect, resulting from an error in applying the chain rule such as $g_x=f_u+f_v$. Distractor C is incorrect, resulting from a sign error such as miscalculating $\partial v / \partial y = -1$. Distractor D is incorrect, corresponding to the value of $f_{uv}(0,0)$.
9. The temperature $u(x, t)$ in a one-dimensional rod can be modeled by the heat equation $u_t = k u_{xx}$ for some constant $k$. For which value of $k$ is $u(x, t) = e^{-4t} \cos(\frac{x}{2})$ a solution to this equation?
   Answer: $16$ When you encounter a partial differential equation like the heat equation and need to verify if a given function is a solution, you must substitute the function into the equation and check if both sides are equal. The heat equation is $u_t = k u_{xx}$, so you need to find the partial derivatives of $u(x,t) = e^{-4t}\cos(\frac{x}{2})$. First, find $u_t$ by differentiating with respect to $t$: $u_t = \frac{\partial}{\partial t}[e^{-4t}\cos(\frac{x}{2})] = -4e^{-4t}\cos(\frac{x}{2})$ Next, find $u_{xx}$ by differentiating twice with respect to $x$: $u_x = \frac{\partial}{\partial x}[e^{-4t}\cos(\frac{x}{2})] = e^{-4t} \cdot (-\sin(\frac{x}{2})) \cdot \frac{1}{2} = -\frac{1}{2}e^{-4t}\sin(\frac{x}{2})$ $u_{xx} = \frac{\partial}{\partial x}[-\frac{1}{2}e^{-4t}\sin(\frac{x}{2})] = -\frac{1}{2}e^{-4t} \cdot \cos(\frac{x}{2}) \cdot \frac{1}{2} = -\frac{1}{4}e^{-4t}\cos(\frac{x}{2})$ Substituting into the heat equation: $-4e^{-4t}\cos(\frac{x}{2}) = k \cdot (-\frac{1}{4}e^{-4t}\cos(\frac{x}{2}))$ Solving for $k$: $-4 = k \cdot (-\frac{1}{4})$, so $k = 16$. This confirms answer (D). Looking at the wrong answers: (A) gives $k = 4$, which would require $u_{xx} = -e^{-4t}\cos(\frac{x}{2})$. (B) gives $k = -16$, which would make the right side positive. (C) gives $k = 1$, requiring $u_{xx} = -4e^{-4t}\cos(\frac{x}{2})$. Remember: always compute both required partial derivatives carefully, paying close attention to the chain rule when differentiating composite functions.
10. Consider the function $h(x,y) = x^3 + y^3 - 3axy$ where $a > 0$. At the critical point $(a,a)$, what is the value of $\frac{\partial^2 h}{\partial x \partial y}$?
    Answer: $-3a$ When you encounter mixed partial derivatives in multivariable calculus, remember that you're looking at how the rate of change in one variable depends on changes in another variable. The notation $\frac{\partial^2 h}{\partial x \partial y}$ means you first take the partial derivative with respect to $y$, then with respect to $x$. Let's work through this systematically. Starting with $h(x,y) = x^3 + y^3 - 3axy$, first find $\frac{\partial h}{\partial y}$: $\frac{\partial h}{\partial y} = 3y^2 - 3ax$ Now take the partial derivative of this result with respect to $x$: $\frac{\partial^2 h}{\partial x \partial y} = \frac{\partial}{\partial x}(3y^2 - 3ax) = -3a$ Notice that this mixed partial derivative is constant — it doesn't depend on the values of $x$ or $y$, so it has the same value at every point, including the critical point $(a,a)$. Looking at the wrong answers: Choice A ($3a$) has the wrong sign — you might get this if you mistakenly wrote $+3ax$ instead of $-3ax$ in the original function. Choice B ($-3$) omits the parameter $a$ entirely, suggesting confusion about which terms contribute to the mixed partial. Choice C ($3$) combines both errors — wrong sign and missing the parameter $a$. Study tip: For mixed partials, work step-by-step and pay careful attention to signs and coefficients. Also remember that for functions with continuous second derivatives, $\frac{\partial^2 h}{\partial x \partial y} = \frac{\partial^2 h}{\partial y \partial x}$ — you can verify your answer by computing it both ways.
11. If $h(x,y) = e^{xy} \sin(x+y)$ and $\frac{\partial h}{\partial x}(0,\pi) = a$, what is $\frac{\partial h}{\partial y}(\pi,0)$?
    Answer: $a$ Using the product rule: $\frac{\partial h}{\partial x} = ye^{xy}\sin(x+y) + e^{xy}\cos(x+y)$. At $(0,\pi)$: $a = \pi e^0 \sin(\pi) + e^0 \cos(\pi) = \pi(0) + 1(-1) = -1$. For $\frac{\partial h}{\partial y} = xe^{xy}\sin(x+y) + e^{xy}\cos(x+y)$. At $(\pi,0)$: $\frac{\partial h}{\partial y}(\pi,0) = \pi e^0 \sin(\pi) + e^0 \cos(\pi) = \pi(0) + 1(-1) = -1 = a$.
12. The ideal gas law can be written as $P = \frac{nRT}{V}$, where $n$ and $R$ are constants. This law defines relationships between pressure ($P$), volume ($V$), and temperature ($T$). What is the value of the product $\left(\frac{\partial P}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P \left(\frac{\partial T}{\partial P}\right)_V$?
    Answer: $-1$ When you encounter partial derivatives with subscripts in thermodynamics, you're dealing with a fundamental relationship in multivariable calculus. The subscripts indicate which variable is held constant during differentiation, and this product represents a cyclic chain of partial derivatives. To find each partial derivative from the ideal gas law $P = \frac{nRT}{V}$, treat the subscripted variable as constant. First, $\left(\frac{\partial P}{\partial V}\right)_T = -\frac{nRT}{V^2}$ since pressure decreases as volume increases at constant temperature. For the second term, rearrange to solve for $V$: $V = \frac{nRT}{P}$, so $\left(\frac{\partial V}{\partial T}\right)_P = \frac{nR}{P}$. For the third term, rearrange to solve for $T$: $T = \frac{PV}{nR}$, so $\left(\frac{\partial T}{\partial P}\right)_V = \frac{V}{nR}$. Now multiply all three: $\left(-\frac{nRT}{V^2}\right) \cdot \left(\frac{nR}{P}\right) \cdot \left(\frac{V}{nR}\right) = -\frac{(nR)^2TV}{V^2 \cdot P \cdot nR} = -\frac{nRT}{VP}$. Since $P = \frac{nRT}{V}$, we have $\frac{nRT}{VP} = 1$, making our product equal to $-1$. Choice (A) gives 1, which ignores the negative sign from the inverse relationship between pressure and volume. Choice (C) suggests no relationship exists, which is impossible for related thermodynamic variables. Choice (D) incorrectly assumes the constants don't cancel out. Remember: cyclic partial derivative products for three interdependent variables always equal $-1$ – this is a universal theorem in multivariable calculus that applies beyond just thermodynamics.
13. If $z = e^{x+y}\cos(x-y)$ and $\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = ke^{x+y}\cos(x-y)$ for some constant $k$, what is the value of $k$?
    Answer: $2$ Using the product rule: $\frac{\partial z}{\partial x} = e^{x+y}\cos(x-y) + e^{x+y}(-\sin(x-y)) = e^{x+y}[\cos(x-y) - \sin(x-y)]$. Similarly: $\frac{\partial z}{\partial y} = e^{x+y}\cos(x-y) + e^{x+y}(-\sin(x-y))(-1) = e^{x+y}[\cos(x-y) + \sin(x-y)]$. Adding these: $\frac{\partial z}{\partial x} + \frac{\partial z}{\partial y} = e^{x+y}[\cos(x-y) - \sin(x-y) + \cos(x-y) + \sin(x-y)] = e^{x+y}[2\cos(x-y)] = 2e^{x+y}\cos(x-y)$. Therefore, $k = 2$.
14. Let $f(x, y) = \ln(x + e^{2y})$. What is the value of the third-order partial derivative $f_{xxy}(0, 0)$?
    Answer: 4 We compute the partial derivatives sequentially: $f_x$, then $f_{xx}$, then $f_{xxy}$. 1. Find $f_x$: $f_x = \frac{\partial}{\partial x} \ln(x + e^{2y}) = \frac{1}{x + e^{2y}}$. 2. Find $f_{xx}$: $f_{xx} = \frac{\partial}{\partial x} (x + e^{2y})^{-1} = -1(x + e^{2y})^{-2} \cdot 1 = -\frac{1}{(x + e^{2y})^2}$. 3. Find $f_{xxy}$: $f_{xxy} = \frac{\partial}{\partial y} \left( -(x + e^{2y})^{-2} \right)$. Using the chain rule: $f_{xxy} = -(-2)(x + e^{2y})^{-3} \cdot \frac{\partial}{\partial y}(x + e^{2y}) = 2(x + e^{2y})^{-3} \cdot (2e^{2y}) = \frac{4e^{2y}}{(x + e^{2y})^3}$. 4. Evaluate at the point $(0, 0)$: $f_{xxy}(0, 0) = \frac{4e^{2(0)}}{(0 + e^{2(0)})^3} = \frac{4e^0}{(0 + e^0)^3} = \frac{4(1)}{(1)^3} = 4$. Distractor B is incorrect; this results from forgetting the factor of 2 from the chain rule when differentiating $e^{2y}$ in the final step. Distractor C is incorrect; this results from a sign error, likely when differentiating $(x+e^{2y})^{-2}$. Distractor D is incorrect; this could result from treating $e^{2y}$ as a constant with respect to $y$.

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