Partial Derivatives

$f_x = 3x^2y - 4y^2$ (the term $2y$ is a constant in $x$).

$f_y = x^3 - 8xy + 2$.

$f_x(1, -1) = -3 - 4 = -7$; $\ f_y(1, -1) = 1 + 8 + 2 = 11$.

$f_x = \cos(xy^2) \cdot y^2 = y^2\cos(xy^2)$.

$f_y = \cos(xy^2) \cdot 2xy = 2xy\cos(xy^2)$.

$\cos(\pi) = -1$: $\ f_x(\pi, 1) = -1$, $\ f_y(\pi, 1) = -2\pi$.

$f_x = \dfrac{z}{y + z}$ (the fraction is $x$ times a constant).

$f_y = -\dfrac{xz}{(y + z)^2}$; quotient rule in $z$: $f_z = \dfrac{x(y + z) - xz}{(y + z)^2} = \dfrac{xy}{(y + z)^2}$.

$f_x = \tfrac{1}{2}$, $\ f_y = -\tfrac{2}{4} = -\tfrac{1}{2}$, $\ f_z = \tfrac{2}{4} = \tfrac{1}{2}$.

$f_x(0, 0) = \displaystyle\lim_{h\to 0}\dfrac{f(h, 0) - f(0, 0)}{h}$, $\ f_y(0, 0) = \displaystyle\lim_{h\to 0}\dfrac{f(0, h) - f(0, 0)}{h}$.

$f(0, 0) = 0$.

$f(h, 0) = \dfrac{h^3}{h^2} = h$ and $f(0, h) = \dfrac{2h^3}{h^2} = 2h$.

$f_x(0, 0) = \displaystyle\lim_{h\to 0}\dfrac{h}{h} = 1$, $\ f_y(0, 0) = \displaystyle\lim_{h\to 0}\dfrac{2h}{h} = 2$.

$f_x(0, 0) = \displaystyle\lim_{h\to 0}\dfrac{f(h, 0) - f(0, 0)}{h}$, and the same with $f(0, h)$ for $f_y$.

$f(0, 0) = 0$.

$f(h, 0) = \dfrac{0}{h^2} = 0$ and $f(0, h) = 0$.

Both quotients are $\dfrac{0}{h} = 0$, so both limits are $0$. (Yet $f$ is not continuous at the origin: along $y = x$, $f = \tfrac{5}{2}$.)

$f_x(0, 0) = \displaystyle\lim_{h\to 0}\dfrac{f(h, 0) - f(0, 0)}{h}$.

$f(0, 0) = 0$.

$f(h, 0) = \sqrt{h^2} = |h|$.

$\dfrac{|h|}{h}$ is $1$ for $h > 0$ and $-1$ for $h < 0$: the one-sided limits differ.

$1 \cdot 2 + 1 \cdot 8 = 10$. ✓

In $x$: $\ 2xz + x^2 z_x + 3yz^2 z_x = 0$.

$z_x = -\dfrac{2xz}{x^2 + 3yz^2}$. In $y$: $x^2z_y + z^3 + 3yz^2z_y = 0$, so $z_y = -\dfrac{z^3}{x^2 + 3yz^2}$.

$x^2 + 3yz^2 = 1 + 12 = 13$: $\ z_x = -\tfrac{4}{13}$, $\ z_y = -\tfrac{8}{13}$.

$e^0 + 1 = 2$. ✓

In $x$ (product rule in the exponent): $\ e^{xz}\left(z + xz_x\right) + yz_x = 0$.

$z_x = -\dfrac{z\,e^{xz}}{x\,e^{xz} + y}$. In $y$: $e^{xz}\,x z_y + z + yz_y = 0$, so $z_y = -\dfrac{z}{x\,e^{xz} + y}$.

$x\,e^{xz} + y = 0 + 1 = 1$: $\ z_x = -1$, $\ z_y = -1$.

$2 + 6 + 3 = 11$. ✓

Take $F = xy + yz + xz - 11$: $\ F_x = y + z$, $\ F_y = x + z$, $\ F_z = x + y$.

$z_x = -\dfrac{y + z}{x + y}$, $\ z_y = -\dfrac{x + z}{x + y}$.

$z_x = -\tfrac{5}{3}$, $\ z_y = -\tfrac{4}{3}$.

Four of them: $f_{xx}$, $f_{xy}$, $f_{yx}$, $f_{yy}$.

Clairaut: $f_{yx} = f_{xy}$, so only three need computing.

$f_x = 2xe^{3y}$, $\ f_y = 3x^2e^{3y}$; then $f_{xx} = 2e^{3y}$, $\ f_{xy} = 6xe^{3y}$, $\ f_{yy} = 9x^2e^{3y}$.

Already simple.

$f_{yxx}$: $y$ first, then $x$ twice.

$f_{yxx} = f_{xxy}$. Taking $x$ first wipes out the messy term, which has no $x$.

$f_x = 3x^2y^2$, $\ f_{xx} = 6xy^2$, $\ f_{xxy} = 12xy$.

Already simple.

$\dfrac{\partial^2 f}{\partial x\,\partial y}$: read right to left, so $y$ first, then $x$.

Either order gives the same answer; $y$ first is fine.

$f_y = \dfrac{2}{x + 2y}$, then $f_{yx} = -\dfrac{2}{(x + 2y)^2}$.

At $(1, 0)$: $-\dfrac{2}{1} = -2$.

We would need $f_{xy} = (f_x)_y$ and $f_{yx} = (f_y)_x$.

If $f$ existed, its mixed partials would be continuous here, so Clairaut gives $f_{xy} = f_{yx}$.

$(f_x)_y = x$ and $(f_y)_x = 2x$.

$x \ne 2x$ when $x \ne 0$, so the two mixed partials cannot match.

$u_{xx}$ and $u_{yy}$.

$u_x = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}$, $\ u_{xx} = \dfrac{2x^3 - 6xy^2}{(x^2 + y^2)^3}$; $\ u_y = -\dfrac{2xy}{(x^2 + y^2)^2}$, $\ u_{yy} = \dfrac{6xy^2 - 2x^3}{(x^2 + y^2)^3}$.

$u_{xx} + u_{yy} = \dfrac{2x^3 - 6xy^2 + 6xy^2 - 2x^3}{(x^2 + y^2)^3} = 0$.

It holds at every $(x, y) \ne (0, 0)$, so $u$ is harmonic.

$u_t$ and $u_{xx}$.

$u_t = -k\,e^{-kt}\sin(3x)$, $\ u_{xx} = -9\,e^{-kt}\sin(3x)$.

$-k\,e^{-kt}\sin(3x) = -18\,e^{-kt}\sin(3x)$.

Cancel $e^{-kt}$ and compare where $\sin(3x) \ne 0$: $-k = -18$. (Where $\sin(3x) = 0$, both sides are $0$.)

$u_{tt}$ and $u_{xx}$.

$u_t = 12(x + 4t)^2$, $\ u_{tt} = 96(x + 4t)$; $\ u_x = 3(x + 4t)^2$, $\ u_{xx} = 6(x + 4t)$.

$96(x + 4t) = 6c^2(x + 4t)$.

For all $x, t$: $96 = 6c^2$, so $c^2 = 16$.