Differentiating the Normal Distribution

168. Differentiating the Normal Distribution

𝑓 (𝑥) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})

We want $𝑓^{'} (𝑥) = \frac{d}{d 𝑥} 𝑓 (𝑥)$

Step 1. Structure

Two factors:

𝑓 (𝑥) = \underset{constant}{\underset{⏟}{\frac{1}{\sqrt{2 𝜋 𝜎^{2}}}}} \cdot \underset{function of 𝑥}{\underset{⏟}{exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})}}

First factor is a constant (does not depend on $𝑥$ ):

𝑓^{'} (𝑥) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot \frac{d}{d 𝑥} exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})

Step 2: Set up the chain rule

𝑢 (𝑥) = - \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}

Then

exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}) = exp (𝑢 (𝑥))

The chain rule:

\frac{d}{d 𝑥} exp (𝑢 (𝑥)) = exp (𝑢 (𝑥)) \cdot 𝑢^{'} (𝑥)

We need 2 things:

Step 3. Differentiate $𝑢 (𝑥)$

𝑢 (𝑥) = - \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}

3.a. Pull out constant

𝑢 (𝑥) = - \frac{1}{2 𝜎^{2}} \cdot {(𝑥 - 𝜇)}^{2}

The factor $- \frac{1}{2} 𝜎^{2}$ depends only on $𝜎$ , not $𝑥$ . Constants pull out of derivatives:

𝑢^{'} (𝑥) = - \frac{1}{2 𝜎^{2}} \cdot \frac{d}{d 𝑥} {(𝑥 - 𝜇)}^{2}

Now we need $\frac{d}{d 𝑥} {(𝑥 - 𝜇)}^{2}$

3.b. Apply chain rule again to ${(𝑥 - 𝜇)}^{2}$

This is itself a composition:

Chain rule:

\frac{d}{d 𝑥} {(𝑥 - 𝜇)}^{2} = \frac{d}{d 𝑣} (𝑣^{2}) \cdot \frac{d 𝑣}{d 𝑥}

Outer derivative:

\frac{d}{d 𝑣} (𝑣^{2}) = 2 𝑣 = 2 (𝑥 - 𝜇)

Inner derivative: $𝑣 (𝑥) = 𝑥 - 𝜇$

Since $𝜇$ is a constant, $\frac{d 𝑢}{d 𝑥} = 0$ . So:

\frac{d 𝑣}{d 𝑥} = \frac{d}{d 𝑥} (𝑥 - 𝜇) = \frac{d}{d 𝑥} (𝑥) - \frac{d}{d 𝑥} (𝜇) = 1 - 0 = 1

Combine:

\frac{d}{d 𝑥} {(𝑥 - 𝜇)}^{2} = 2 (𝑥 - 𝜇) \cdot 1 = 2 (𝑥 - 𝜇)

3.c. Put Step 3 together

Substitute back into expression for $𝑢^{'} (𝑥)$

𝑢^{'} (𝑥) = - \frac{1}{2 𝜎^{2}} \cdot (𝑥 - 𝜇)

Simplify be canceling the 2′s:

𝑢^{'} (𝑥) = - (2 \frac{𝑥 - 𝜇}{2 𝜎^{2}}) = - \frac{𝑥 - 𝜇}{𝜎^{2}}

So:

𝑢^{'} (𝑥) = - \frac{𝑥 - 𝜇}{𝜎^{2}}

Step 4. Apply chain tule from Step 2

We had:

\frac{d}{d 𝑥} exp (𝑢 (𝑥)) = exp (𝑢 (𝑥)) \cdot 𝑢^{'} (𝑥)

Plug in:

So:

\frac{d}{d 𝑥} exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}) = exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}) \cdot (- \frac{𝑥 - 𝜇}{𝜎^{2}})

Move the factor to the front for clarity:

- \frac{𝑥 - 𝜇}{𝜎^{2}} \cdot exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})

Step 5. Putting everything together

Recall from Step 1:

𝑓^{'} (𝑥) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot \frac{d}{d 𝑥} exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})

Substitute what we found:

𝑓^{'} (𝑥) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot [- \frac{𝑥 - 𝜇}{𝜎^{2}} \cdot exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})]

Pull the $- \frac{𝑥 - 𝜇}{𝜎^{2}}$ out front:

𝑓^{'} (𝑥) = - \frac{𝑥 - 𝜇}{𝜎^{2}} \cdot \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})

Step 6. Recognize $𝑓 (𝑥)$ on the right side

Look at the last two factors

\frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}) = 𝑓 (𝑥)

That’s exactly the original PDF. So:

𝑓^{'} (𝑥) = - \frac{𝑥 - 𝜇}{𝜎^{2}} \cdot 𝑓 (𝑥)

Step 7. Standard Normal

Does this collapse to $𝜙^{'} (𝑧)$ when $𝜇 = 0$ and $𝜎 = 1$ ?

Plug in $𝜇 = 0$ , $𝜎 = 1$ :

𝑓^{'} (𝑥) = - \frac{𝑥 - 0}{1^{2}} \cdot 𝑓 (𝑥) = - 𝑥 \cdot 𝑓 (𝑥)

With $𝜇 = 0$ and $𝜎 = 1$ , $𝑓 (𝑥)$ becomes $𝜙 (𝑥)$ , so:

𝜙 (𝑥) = - 𝑥 \cdot 𝜙 (𝑥)