Higher-Order & Mixed Partials - Calculus 3

Given $f=x^3y^2+y\ln x$ , find $f_{xyx}$ at $(1,2)$ .
Answer: 23 Differentiate with respect to x first: $f_x$ = $3x^2$ $y^2$ + y/x. Then with respect to y: f_xy = $6x^2$ y + 1/x. Differentiate that with respect to x: f_xyx = 12xy - 1/ $x^2$ . At (1,2), this is 12(1)(2) - 1 = 23. The tempting wrong answer 24 comes from dropping the -1/ $x^2$ term.
For smooth $f,g=f(x^2-y,xy)$ . At $(1,0)$ : $f_u=3,f_v=-2$ , all second partials of $f$ equal 1. Find $g_{xy}$ .
Answer: $g_{xy}=-2$ Using the chain rule, $g_x$ = 2x $f_u$ + y $f_v$ . Differentiate that with respect to y: g_xy = 2x(f_uu $u_y$ + f_uv $v_y$ ) + $f_v$ + y(f_vu $u_y$ + f_vv $v_y$ ). At (1,0), $u_y$ = -1 and $v_y$ = 1, and all second partials are 1, so both derivative-of-f terms vanish. Only $f_v$ = -2 remains, giving g_xy = -2. The tempting mistake is dropping the $f_v$ term when differentiating y $f_v$ .
For $r(x,y) = \arctan\left(\frac{y}{x}\right)$ where $x > 0$ , if $r_{xy}(a,b) = \frac{1}{5}$ for some point $(a,b)$ , what is $a^2 + b^2$ ?
Answer: $5$ When you encounter mixed partial derivatives, you're working with how a function changes when you vary one variable, then another. This tests your ability to apply the chain rule and quotient rule systematically. To find $r_{xy}$ , you need to compute $\frac{\partial}{\partial y}\left(\frac{\partial r}{\partial x}\right)$ . First, find $r_x$ : $r_x = \frac{\partial}{\partial x}\arctan\left(\frac{y}{x}\right) = \frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot \frac{-y}{x^2} = \frac{-y}{x^2 + y^2}$ Next, differentiate with respect to $y$ : $r_{xy} = \frac{\partial}{\partial y}\left(\frac{-y}{x^2 + y^2}\right)$ Using the quotient rule: $\frac{d}{dy}\left(\frac{-y}{x^2 + y^2}\right) = \frac{-(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} = \frac{-x^2 + y^2}{(x^2 + y^2)^2}$ Setting this equal to $\frac{1}{5}$ : $\frac{y^2 - x^2}{(x^2 + y^2)^2} = \frac{1}{5}$ Cross-multiplying: $5(y^2 - x^2) = (x^2 + y^2)^2$ Let $s = x^2 + y^2$ . Then $y^2 - x^2 = 2y^2 - s$ , so: $5(2y^2 - s) = s^2$ $10y^2 - 5s = s^2$ $s^2 + 5s - 10y^2 = 0$ Since this must hold and we need a specific value, testing $s = 5$ gives us a consistent solution. Choice A (50) likely comes from squaring an intermediate result incorrectly. Choice B (10) might result from confusing the coefficient in our equation. Choice C (25) could come from squaring 5 prematurely. Strategy tip: With mixed partials involving inverse trig functions, always apply the quotient rule carefully and substitute systematically to avoid algebraic errors.
Let $f(x,y)=\int_0^{xy} e^{-t^2}dt$ . Find $f_{xy}(1,1)$ .
Answer: $-e^{-1}$ Differentiate with respect to x: $f_x$ = y $e^{-(xy)^2}$ . Then differentiate with respect to y using the product rule: f_xy = $e^{-(xy)^2}$ + y $e^{-(xy)^2}$ ( $-2x^2$ y). At (1,1) this is $e^{-1}$ (1 - 2) = - $e^{-1}$ . A tempting mistake is $e^{-1}$ , which comes from keeping only the first product-rule term and missing the chain-rule factor - $2x^2$ y.
If $f_{xy}(0,0)=5$ and $f_{yx}(0,0)=-5$ , then necessarily...
Answer: A mixed partial discontinuous If both mixed partials f_xy and f_yx were continuous at the origin, they would have to be equal by Clairaut's theorem. Since they are 5 and -5, at least one of those mixed partials must fail to be continuous there. The tempting wrong answer is that f itself must be discontinuous; in fact, f can still be continuous at the origin while its mixed partials differ.
Let $f(x,y) = x^3y^2 + 2xy^4 - 3x^2y$ . If $f_{xxy}(a,b) = 12b - 6$ for some point $(a,b)$ , what is the value of $f_{yxx}(a,b)$ ?
Answer: $12b - 6$ By Clairaut's theorem, mixed partial derivatives are equal when the function has continuous second derivatives, so $f_{xxy} = f_{yxx}$ . Since $f_{xxy}(a,b) = 12b - 6$ , we have $f_{yxx}(a,b) = 12b - 6$ . Choice B incorrectly substitutes y for b. Choice C attempts to compute a different mixed partial. Choice D represents a computational error in finding the partial derivatives.
For the function $v(x,y) = x^2e^y + ye^x$ , which relationship holds between the mixed partial derivatives?
Answer: $v_{xy} = v_{yx} = 2xe^y + e^x$ Computing $v_x = 2xe^y + ye^x$ and $v_y = x^2e^y + e^x$ . Then $v_{xy} = \frac{\partial}{\partial y}(2xe^y + ye^x) = 2xe^y + e^x$ and $v_{yx} = \frac{\partial}{\partial x}(x^2e^y + e^x) = 2xe^y + e^x$ . By Clairaut's theorem, these are equal since $v$ has continuous second partial derivatives. Choice B incorrectly suggests they're different. Choice C has the wrong expression. Choice D includes an extra term.
At $(0,0)$ set $f=0$ ; elsewhere $f=xy(x^2-y^2)/(x^2+y^2)$ . Which is true?
Answer: $f_{xy}(0,0)=-f_{yx}(0,0)$ Compute along the axes: $f_x$ (0,k) = -k and $f_y$ (h,0) = h, while both first partials at (0,0) are 0. Therefore f_xy(0,0) = lim (-k)/k = -1 and f_yx(0,0) = lim h/h = 1. The mixed partials are negatives of each other. The tempting wrong answer is that both are 0 because f vanishes on the axes, but the partial derivatives along the axes are not 0 away from the origin.
If $w(x,y) = \ln(x^2 + y^2)$ and both $w_{xx}(a,b)$ and $w_{yy}(a,b)$ exist, what is $w_{xx}(a,b) + w_{yy}(a,b)$ ?
Answer: $0$ First, $w_x = \frac{2x}{x^2 + y^2}$ and $w_y = \frac{2y}{x^2 + y^2}$ . Then $w_{xx} = \frac{2(x^2 + y^2) - 2x(2x)}{(x^2 + y^2)^2} = \frac{2y^2 - 2x^2}{(x^2 + y^2)^2}$ and $w_{yy} = \frac{2(x^2 + y^2) - 2y(2y)}{(x^2 + y^2)^2} = \frac{2x^2 - 2y^2}{(x^2 + y^2)^2}$ . Therefore, $w_{xx} + w_{yy} = \frac{2y^2 - 2x^2 + 2x^2 - 2y^2}{(x^2 + y^2)^2} = 0$ . This shows that $w$ is harmonic. Choices A and B give incorrect partial sums. Choice D incorrectly simplifies the arithmetic.
Consider $p(x,y,z) = xyz + x^2z^2 + y^3z$ . What is the value of $p_{xyz} + p_{zxy} + p_{yzx}$ ?
Answer: $3$ By Clairaut's theorem, all mixed third partial derivatives with the same variables are equal: $p_{xyz} = p_{zxy} = p_{yzx}$ . Computing $p_{xyz}$ : $p_x = yz + 2xz^2$ , $p_{xy} = z$ , and $p_{xyz} = 1$ . Therefore, $p_{xyz} + p_{zxy} + p_{yzx} = 1 + 1 + 1 = 3$ . Choice A gives only one term. Choice C incorrectly multiplies by 6. Choice D represents $3^2$ instead of $3 \times 1$ .

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