diff --git a/documents/by-course/math-6a/course-notes/main.typ b/documents/by-course/math-6a/course-notes/main.typ
index 69d4719..25e2415 100644
--- a/documents/by-course/math-6a/course-notes/main.typ
+++ b/documents/by-course/math-6a/course-notes/main.typ
@@ -772,5 +772,309 @@ $
   (dif) / (dif x) cosh(x) = sinh(x)
 $
 
-It's pretty easy to show these using their definitions, and derive the
-derivative of $tanh$.
+It's pretty easy to show these using their definitions, and derive the derivative of $tanh$.
+
+== End of weeks 1-4
+
+That was all of the content of week 1 to 4. Now we shift to weeks 5-7, where we studied more about vectors and their derivatives.
+
+== Partial derivatives
+
+These are the slopes of tangent lines for the graph in the direction of the
+changing variable.
+
+#theorem[Clairaut's theorem][
+  Suppose $f : RR^2 -> RR$ is defined on a disk $D$ that contains a point
+  $(a,b)$. If the functions $f_(x y)$ and $f_(y x)$ are ocntinuous on this disk
+  then they are the same.
+]
+
+#theorem[Extended Clairaut][
+  Suppose $f : RR^2 -> RR$ is defined on a disk $D$ that contains $(a,b)$. If
+  all of the mixed partial derivatives are continuous anywhere in the disk $D$,
+  then the mixed partials are equal.
+]
+
+== Multivariable chain rule
+
+The product rule actually follows from it. Recall:
+$
+  (f(x) g(x))' = f'(x) g(x) + f(x) g'(x)
+$
+Then instead let's replace $f$ and $g$ with $x$ and $y$, such that we have something like
+$
+  z = x y
+$
+Then let $x$ and $y$ be functions of $t$. So
+$
+  z = x(t) y(t)
+$
+The partial derivatives
+$
+  (diff z) / (diff x) = y, (diff z) / (diff y) = x
+$
+By the multivariable chain rule,
+$
+  (diff z) / (diff t) = x'(t) (diff z) / (diff x) + (diff z) / (diff y) y'(t) = x'(t) y(t) + x(t) y'(t)
+$
+
+== Implicit differentiation
+
+It's similar to the single variable implicit differentiation, but remember to
+hold the extraneous variables constant in practice.
+
+#example[
+  Suppose you have a surface
+  $
+    3x^2 + 5 y z + z^3 = 0
+  $
+  And you want a partial derivative $(diff y)/(diff z)$ at some point. You can
+  use implicit differentiation by viewing the surface as a level set of some
+  larger function $F(x,y,z) = 3x^2 + 5y z + z^3$ where $F(x,y,z) = 0$.
+
+  Now we differentiate both sides:
+  $
+    (diff F) / (diff z) = diff / (diff z)(3x^2) + diff / (diff z)(5y z) + diff / (diff z)(z^3) \
+    = 0 + (5 (diff y) / (diff z) + 5y) + 3z^2
+  $
+  Then solve for $(diff y)/(diff z)$.
+]
+
+== Multivariable chain rule as matrix
+
+Consider $z = f(x,y)$. Then the derivative of $z$ with respect to $x$ and $y$ would be a matrix:
+$
+  mat((diff z)/(diff x), (diff z)/(diff y))
+$
+
+Now suppose the coordinate system changes to
+$
+  x = 3u - v \
+  y = 2v
+$
+Now suppose we want $z_u$ and $z_v$ at $(x,y) = (3,6)$. Then this is actually
+$(u,v) = (2,3)$ in $u v$ coordinates. The partials of $z$ with respect to $u$
+and $v$ are just matrix multiplication:
+$
+  mat(z_u,z_v) = mat(z_x, z_y) mat(x_u,x_v;y_u,y_v)
+$
+
+== Differentials
+
+Differentials are about linear approximation. Recall in the single variable
+case we use the tangent line approximation and differentials to approximate
+functions. In the multivariable case it's the tangent plane approximation and
+the directional derivative.
+
+Let $f$ be a function with two inputs and one output, say
+$
+  f(x,y) = x^2 + x cos(y)
+$
+
+Then the tangent plane at $(x_0,y_0)$ is the plane that best approximates the
+function at that point. Now we need two slopes instead of one.
+
+Take a look at $(1,pi/2)$. Then the partial derivatives at that point are 2 and
+-1. The idea is we start at $(1,pi/2,1)$ and then move a slight nudge in either
+the $x$ or $y$ directions. In the $x$ directions, we move by $Delta x$ and get
+an increase in $z$ (height) of $2 dot Delta x$.
+
+If we move a slight nudge in the $y$ direction $Delta y$, then our height
+should increase (decrease) by $-1 dot Delta y$.
+
+Then we have a tangent plane approximation of
+$
+  Delta z approx 2 dot Delta x - 1 dot Delta y
+$
+
+And like in single variable, we can replace the $Delta$ with $dif$.
+$
+  diff z approx 2 dot diff x - 1 dot diff y
+$
+Sidenote: how can we actually have the equation of the tangent plane in terms
+of $x$,$y$,$z$? Just note that $x = 1 + Delta x$, $y = pi/2 + Delta y$, and $z
+= 1 + Delta z$. So just substitute
+$
+  (z - 1) = 2(x-1) - 1(y-pi / 2)
+$
+
+So in general, the differential version
+$
+  dif z = m_x dif x + m_y dif y
+$
+
+and the tangent plane equation at $(x_0, y_0, z_0)$:
+$
+  z = z_0 + m_x (x-x_0) + m_y (y-y_0)
+$
+
+== Directional derivative
+
+Now take another look at the linear approximation in $z$
+$
+  Delta z approx m_x Delta x + m_y Delta y
+$
+
+The directional derivative reinterprets this as matrix multiplication
+$
+  Delta z approx mat(m_x,m_y) vec(Delta x, Delta y)
+$
+This is the same as a dot product, in fact, the dot product of the gradient
+given by $vec(m_x,m_y)$ and a tiny movement vector.
+
+== Directional derivative
+
+We mentioned this before, now let's discuss in more detail. If you move by
+$Delta x$ and $Delta y$, in $x$ and $y$ directions, then the direction
+derivative computes your change in height on the tangent plane.
+
+Consider the question "What is the derivative of $f$" in the direction of the
+vector $vec(3,4)$?
+
+Now consider $m_x (3) + m_y (4) = 3m_x + 4m_y$. This is almost the answer, but
+really this is the change in $f$ resulting from a movement in the direction of
+$vec(3,4)$. If we want the derivative, we're asking for the slope. We have
+rise, now run is $sqrt(3^2 + 4^2) = 5$, so the answer is $(3m_x + 4m_y)/5$.
+
+A more intuitive way is to consider a unit vector $arrow(u) = 1/lr(|<3,4>|)
+<3,4>$ that points in the same direction. So now the "run" is simply 1. Clearly
+we get the same answer, but we have a good formula now
+
+$
+  "Direction derivative" = nabla arrow(f) dot arrow(u)
+$
+
+where $arrow(u)$ is the *unit vector* in our desired direction.
+
+A geometric interpretation is that the direction derivative in a given
+direction is just the gradient vector projected in that direction.
+
+To recap:
+
+We discuss two questions: what is the slope in a given direction, and what
+direction has the steepest slope?
+
+Let $arrow(F)$ be a gradient vector of our partial derivatives and $arrow(v)$
+be the movement vector in the $x y$-plane. Let $arrow(u)$ be a unit vector
+pointing in the same direction.
+
+The answer to the first question is $nabla arrow(f) dot arrow(u)$, where
+$arrow(u)$ is a unit vector in the given direction
+
+The answer to the second question can be derived.
+
+$
+  arrow(F) dot arrow(u) = lr(|arrow(F)|) lr(|arrow(u)|) cos(theta)
+$
+and $theta$ is the angle between $arrow(F)$ and $arrow(u)$. So since $cos
+theta$ reaches maximum value at $theta = 0$, the maximum possible slope is
+actually in the same direction as $arrow(F)$ with a slope equal the magnitude
+of $arrow(F)$.
+
+Takeaways:
+
+- Movement in the direction of the gradient vector gives the "steepest" ascent of the function
+- Movement perpendicular to the gradient has slope of 0
+- Movement in the opposite direction of the gradient has maximum negative slope (sharpest descent) with same magnitude as gradient
+- Any directional derivative in between can be calculated as a projection from the gradient
+
+== Optimization
+
+We spoke previously about the Lagrange multiplier. Now we discuss it in greater
+detail.
+
+When optimizing in two dimensions, we either optimize for all of $RR^2$, or on
+a constraint in $RR^2$ (such as a curve). For the first case, we use critical
+points. For the second, *Lagrange multipliers*.
+
+== Critical points
+
+Critical points occur when the gradient is zero or undefined. Both partials are
+zero *or* at least one of them isn't defined.
+
+Essentially, they occur when the tangent plane is flat. We can't just look at
+$f''(x)$ like in single variable calculus, but we can take the determinant of
+the second order partials for some sort of multivariable concavity. It measures
+how much the pure partial derivatives dominate the mixed partial derivatives,
+and they need to dominant to a certain extent such that there is consistently
+upward or downward curvature in every direction.
+
+We find the critical points when $m_x = 0$ and $m_y = 0$ or either are
+undefined. Then we classify them as follows.
+
+Recall second derivative test for single variable function, now consider the
+two-variable case.
+
+$
+  Dif (x_0,y_0) = det mat(f_(x x) (x_0, y_0), f_(x y) (x_0, y_0); f_(y x) (x_0,y_0), f_(y y) (x_0, y_0)) = f_(x x) (x_0, y_0) f_(y y) (x_0, y_0) - f_(x y) (x_0, y_0)^2
+$
+
+- If $Dif > 0$ and $f_(x x) (x_0, y_0) > 0$, then $f$ is a relative minimum.
+- If $Dif > 0$ and $f_(x x) (x_0, y_0) < 0$, then $f$ is a relative maximum.
+- If $Dif < 0$, then$f(x_0, y_0)$ is neither and it's a saddle point.
+- If $Dif = 0$, then we don't know
+
+== Lagrange multipliers
+
+We discuss optimization on a restricted curve in our domain.
+
+Idea: we should navigate along the curve and find where the direction
+derivative is 0. Recall that this is the same as when the velocity vector is
+perpendicular to the gradient. The issue is that we always have to parametrize
+the curve.
+
+To avoid this, Lagrange multipliers views the (implicit) constraint equation as
+a level curve of another surface. If we take the gradient of that surface
+everywhere on the level curve, then that gradient is parallel to the original
+function's gradient at critical points.
+
+So, we should be able to take the gradients of both the function and the
+constraint function, and look for when one is a scalar multiple of the other.
+
+Consider a function $f$ and a constraint $g$. Then we compute $nabla f$ and
+$nabla g$, then solve for when $nabla f = lambda nabla g$. Then we can just plug in
+points and figure it out.
+
+#exercise[
+  Find the highest and lowest points on $f(x,y) = 81x^2 + y^2$ with the
+  constraint $4x^2 + y^2 = 9$. Let the second function be $g(x,y)$, and keep in
+  mind our constraint is essentially the level set where $g(x,y) = 9$.
+]
+
+Intuition: consider a constraint $g(x,y) = x^2 + y^2$, and our constraint is
+the level set where $g(x,y) = 25$. Notice, the gradient of $g$ is perpendicular
+to its level set at any given point. So, when optimizing on a function $f$ that
+is $g$-constrained, we are really looking for where the gradient of $g$ is
+parallel to the gradient $f$. That is why we are using a scalar multiple
+$lambda$ to relate them.
+
+Computation:
+
+We have $g(x,y) = 25$. We construct the equation
+
+$
+  vec(f_x, f_y) = lambda vec(g_x,g_y)
+$
+
+This gives three equations
+
+$
+  f_x = lambda g_x \
+  f_y = lambda g_y \
+  g(x,y) = 25
+$
+
+We find
+$
+  y = 4 lambda^2 y
+$
+
+If $y != 0$, then $lambda = plus.minus 1/2$. So we're looking for points on the
+circle, with radius 5, such that $x = plus.minus y$. This gives 4 points to
+consider: $(plus.minus 5/sqrt(2), plus.minus 5/sqrt(2))$.
+
+If $y = 0$, then $x$ is forced to be 0, and $(0,0)$ is not on the circle. So we
+ignore it.
+
+Now we just compare our four candidates and find the greatest (or least) for
+optimization!
diff --git a/documents/by-course/math-8/course-notes/main.typ b/documents/by-course/math-8/course-notes/main.typ
index 7c65073..bdf19c2 100644
--- a/documents/by-course/math-8/course-notes/main.typ
+++ b/documents/by-course/math-8/course-notes/main.typ
@@ -920,13 +920,15 @@ nonempty subsets of $A$ whose union is $A$.
 
 == Functions
 
-Let $A$ and $B$ be sets. A relation $R$ from $A$ to $B$ is a subset $R subset.eq A times B$.
-
 #definition[
-  A *function* $f$ from $A$ to $B$ relates each element of $"Dom"(R)$ to to exactly
-  one element of $"Rng"(R)$.
+  A *function* $f$ is a relation from $A$ to $B$ such that
+  1. $"Dom"(f) = A$.
+  2. If $(x,y) in f$ and $(x,z) in f$ then $y = z$.
 ]
 
+That is, every element in $A$ is related to exactly one element in $B$. Note
+that (2) is the vertical line test.
+
 #fact[
   A function from $A$ to $B$ is written
   $