This is a quick math refresh for me to reference in the future - and might be useful to you if you're learning backpropagation:

Example 1

flowchart LR
    a["a<br/>data: 3<br/>grad: 4"] --> plus(("+"))
    b["b<br/>data: 2<br/>grad: 4"] --> plus
    plus --> c["c<br/>data: 5<br/>grad: 4"]
    c --> mult(("×"))
    d["d<br/>data: 4<br/>grad: 5"] --> mult
    mult --> L["L<br/>data: 20<br/>grad: 1"]

gradient of L

Starting at the end, the $\frac{dL}{dL} = 1$ - this is because any change in L produces exactly that same change in L.

gradient of c

Now, on to c. $c*d = L$ . We want to figure out how much a movement in c impacts the result in L.

When you take the derivative of L with respect to c, $\frac{dL}{dc}$ , lets assume the other value $d = 3$ so the value would be $3c = L$ . $\frac{dL}{dc}$ is now just 3 - or d.

Therefore, the gradient of c is the value of d: 4.

gradient of a

Now, to a. $a+b = c$ . Now we want to figure out how much a movement in a impacts the result in L.

First, let's figure out the local derivative of $\frac{dc}{da}$ . Because this is an addition function, the value is 1. Any movement in a will have the exact same movement in c.

Now, we want to figure out:

\frac{dL}{da} = \frac{dL}{dc} * \frac{dc}{da}

This is effectively 4 ( $\frac{dL}{dc}$ )* 1( $\frac{dc}{da}$ ) = 4.

Example 2

a = 2
b = 3
d = 4
f = 2
c = a * b
e = c + d
L = e * f

Forward pass: c = 6, e = 10, L = 20.

flowchart LR
    a["a<br/>data: 2<br/>grad: 6"] --> mult1(("×"))
    b["b<br/>data: 3<br/>grad: 4"] --> mult1
    mult1 --> c["c<br/>data: 6<br/>grad: 2"]
    c --> plus(("+"))
    d["d<br/>data: 4<br/>grad: 2"] --> plus
    plus --> e["e<br/>data: 10<br/>grad: 2"]
    e --> mult2(("×"))
    f["f<br/>data: 2<br/>grad: 10"] --> mult2
    mult2 --> L["L<br/>data: 20<br/>grad: 1"]

gradient of L

Same as last time, the gradient is 1.

gradient of e

$\frac{dL}{de}$ = the value of f, which is 2

gradient of c

Local derivative first: $\frac{de}{dc}$ = 1 because its an addition equation.

\frac{dL}{dc} = \frac{dL}{de} * \frac{de}{dc}

which is = 2 * 1 or 2.

gradient of a

Local derivative first: $\frac{dc}{da}$ = the value of b, which is 3.

\frac{dL}{da} = \frac{dL}{de} * \frac{de}{dc} * \frac{dc}{da}

which is = 2 * 1 * 3 = 6.

Example 3

x = 5
y = x + 2
z = x * 3
L = y * z

Forward pass: y = 7, z = 15, L = 105.

flowchart LR
    x["x<br/>data: 5<br/>grad: 36"] --> plus(("+"))
    two["2<br/>(const)"] --> plus
    plus --> y["y<br/>data: 7<br/>grad: 15"]
    x --> mult1(("×"))
    three["3<br/>(const)"] --> mult1
    mult1 --> z["z<br/>data: 15<br/>grad: 7"]
    y --> mult2(("×"))
    z --> mult2
    mult2 --> L["L<br/>data: 105<br/>grad: 1"]

gradient of L

$\frac{dL}{dL} = 1$

gradient of y

$\frac{dL}{dy} = 15$ (the value of z)

gradient of x

Because x is involved in two different paths, we need to calculate both paths and add them together"

via y:

$\frac{dL}{dx} = \frac{dL}{dy} * \frac{dy}{dx}$

$\frac{dy}{dx}$ is 1, so 15*1 = 15

via z:

$\frac{dL}{dx} = \frac{dL}{dz} * \frac{dz}{dx}$

$\frac{dL}{dz}$ = 7

$\frac{dz}{dx}$ = 3 (remember, this is a multiplication, not addition)

Therefore, $\frac{dL}{dx}$ = (7*3) + 15 = 36

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