Multivariable Chain Rule - Calculus 3

1. A function $w = f(x, y)$ satisfies $\frac{\partial w}{\partial x} = 2xy$ and $\frac{\partial w}{\partial y} = x^2 + 3y^2$. If $x = t^2 + 1$ and $y = 2t - 1$, find $\frac{dw}{dt}$ when $t = 1$.
   Answer: $20$ Using the chain rule: $\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt}$. We have $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 2$. When $t = 1$: $x = 2$, $y = 1$, $\frac{dx}{dt} = 2$. So $\frac{\partial w}{\partial x} = 2(2)(1) = 4$ and $\frac{\partial w}{\partial y} = 2^2 + 3(1)^2 = 7$. Therefore $\frac{dw}{dt} = 4(2) + 7(2) = 8 + 12 = 20$. Choice B results from computing $\frac{\partial w}{\partial y}$ incorrectly as $2^2 + 3(1)^2 = 8$. Choice C comes from using $\frac{dx}{dt} = 1$ instead of $2$. Choice D results from multiple computational errors.
2. If $w = xy^2z^3$ where $x = u^2 + v^2$, $y = uv$, and $z = u - v$, which expression correctly represents $\frac{\partial w}{\partial u}$?
   Answer: $2u(uv)^2(u-v)^3 + (u^2+v^2)(2uv)(u-v)^3 + (u^2+v^2)(uv)^2(3(u-v)^2)$ Using the multivariable chain rule: $\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}$. We have $\frac{\partial w}{\partial x} = y^2z^3$, $\frac{\partial w}{\partial y} = 2xyz^3$, $\frac{\partial w}{\partial z} = 3xy^2z^2$, and $\frac{\partial x}{\partial u} = 2u$, $\frac{\partial y}{\partial u} = v$, $\frac{\partial z}{\partial u} = 1$. Substituting gives choice A. Choice B incorrectly applies the product rule to $\frac{\partial w}{\partial y}\frac{\partial y}{\partial u}$. Choice C incorrectly multiplies by an extra factor of $uv$. Choice D incorrectly computes $\frac{\partial w}{\partial y}\frac{\partial y}{\partial u}$.
3. Consider $z = f(u, v)$ where $u = x + 2y$ and $v = x - y$. If $\frac{\partial z}{\partial x} = 4$ and $\frac{\partial z}{\partial y} = 1$ at a particular point, what is the value of $\frac{\partial f}{\partial u}$ at the corresponding point?
   Answer: $\frac{5}{3}$ We have the system: $\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial f}{\partial u}(1) + \frac{\partial f}{\partial v}(1) = 4$ and $\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial f}{\partial v}\frac{\partial v}{\partial y} = \frac{\partial f}{\partial u}(2) + \frac{\partial f}{\partial v}(-1) = 1$. This gives us $\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} = 4$ and $2\frac{\partial f}{\partial u} - \frac{\partial f}{\partial v} = 1$. Adding these equations: $3\frac{\partial f}{\partial u} = 5$, so $\frac{\partial f}{\partial u} = \frac{5}{3}$. Choice B comes from incorrectly solving the system. Choice C results from sign errors. Choice D comes from confusing partial derivatives.
4. Consider the function $h(x, y) = g(x^2 + y^2, x^2 - y^2)$ where $g$ has continuous second partial derivatives. If $\frac{\partial h}{\partial x} = 4x$ when $x = 1$ and $y = 0$, what is the value of $g_1(1, 1) + g_2(1, 1)$?
   Answer: $2$ Let $u = x^2 + y^2$ and $v = x^2 - y^2$. Then $h(x,y) = g(u,v)$ and $\frac{\partial h}{\partial x} = g_1\frac{\partial u}{\partial x} + g_2\frac{\partial v}{\partial x} = g_1(2x) + g_2(2x) = 2x(g_1 + g_2)$. When $x = 1, y = 0$: $u = 1, v = 1$, and $\frac{\partial h}{\partial x} = 2(1)(g_1(1,1) + g_2(1,1)) = 2(g_1(1,1) + g_2(1,1))$. Given that $\frac{\partial h}{\partial x} = 4x = 4(1) = 4$ at this point, we have $2(g_1(1,1) + g_2(1,1)) = 4$, so $g_1(1,1) + g_2(1,1) = 2$. Choice A comes from forgetting the factor of 2. Choice C results from dividing by 4 instead of 2. Choice D comes from not dividing by the coefficient.

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