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Math  /  Calculus

Question2. Let S(t)=11+e−tS(t)=\frac{1}{1+e^{-t}}. a. Find S′(t)S^{\prime}(t). Show your work completely to justify your response. b. Which of the following equations hold true? Explain your thinking fully. (Note: Only one equation is true.) 1) S′(t)=S(t)S^{\prime}(t)=S(t) 2) S′(t)=(S(f))2S^{\prime}(t)=(S(f))^{2} 3) S′(t)=S(t)(1−S(t))S^{\prime}(t)=S(t)(1-S(t)) 4) S′(t)=−S(−t)S^{\prime}(t)=-S(-t)
Note: The function S(f)S(f) is called the "Sigmoid activation function" and is extremely important in machine learning and artificial intelligence.

Studdy Solution

STEP 1

What is this asking? We're finding the derivative of a function and then figuring out which equation, out of a given set, is true for that derivative. Watch out! Don't forget the chain rule when taking the derivative, and be careful with signs when simplifying!

STEP 2

1. Find the derivative
2. Check the equations

STEP 3

Alright, let's **find** that derivative!
Our function is S(t)=11+e−tS(t) = \frac{1}{1 + e^{-t}}.
We can rewrite this as S(t)=(1+e−t)−1S(t) = (1 + e^{-t})^{-1} to make it easier to differentiate.

STEP 4

Now, we'll use the **chain rule**.
The outer function is something to the power of −1-1, and the inner function is 1+e−t1 + e^{-t}.

STEP 5

The derivative of the outer function (something to the power of −1-1) is −1⋅(something)−2-1 \cdot (\text{something})^{-2}.
The derivative of the inner function (1+e−t1 + e^{-t}) is −e−t-e^{-t}.
Remember, the derivative of ee to the *something* is ee to the *something* times the derivative of the *something*.
Here, the *something* is −t-t, and its derivative is −1-1.

STEP 6

Putting it all together with the chain rule, we get: S′(t)=−1⋅(1+e−t)−2⋅(−e−t)S'(t) = -1 \cdot (1 + e^{-t})^{-2} \cdot (-e^{-t})

STEP 7

Let's **simplify** this!
The two negatives multiply to give us a positive: S′(t)=e−t(1+e−t)2S'(t) = \frac{e^{-t}}{(1 + e^{-t})^{2}} So, S′(t)=e−t(1+e−t)2S'(t) = \frac{e^{-t}}{(1 + e^{-t})^{2}}.
That's our derivative!

STEP 8

Let's see which equation is true.
First, we found S′(t)=e−t(1+e−t)2S'(t) = \frac{e^{-t}}{(1 + e^{-t})^{2}}.
Clearly, this isn't equal to S(t)S(t), which is 11+e−t\frac{1}{1 + e^{-t}}.
So, the first equation, S′(t)=S(t)S'(t) = S(t), is **not true**.

STEP 9

The second equation, S′(t)=(S(t))2S'(t) = (S(t))^{2}, is e−t(1+e−t)2=1(1+e−t)2\frac{e^{-t}}{(1 + e^{-t})^{2}} = \frac{1}{(1 + e^{-t})^{2}}.
This is also **not true**.

STEP 10

Now, for the third equation, S′(t)=S(t)(1−S(t))S'(t) = S(t)(1 - S(t)).
Let's **substitute** S(t)S(t) into the right side: S(t)(1−S(t))=11+e−t⋅(1−11+e−t)S(t)(1 - S(t)) = \frac{1}{1 + e^{-t}} \cdot \left( 1 - \frac{1}{1 + e^{-t}} \right)

STEP 11

Let's **simplify** this.
We can rewrite the term in the parentheses by multiplying 11 by 1+e−t1+e−t\frac{1 + e^{-t}}{1 + e^{-t}} to get a common denominator: =11+e−t⋅(1+e−t1+e−t−11+e−t)=11+e−t⋅(e−t1+e−t)= \frac{1}{1 + e^{-t}} \cdot \left( \frac{1 + e^{-t}}{1 + e^{-t}} - \frac{1}{1 + e^{-t}} \right) = \frac{1}{1 + e^{-t}} \cdot \left( \frac{e^{-t}}{1 + e^{-t}} \right)

STEP 12

Multiplying the fractions gives us: e−t(1+e−t)2\frac{e^{-t}}{(1 + e^{-t})^{2}} This is the same as our S′(t)S'(t)!
So, the third equation, S′(t)=S(t)(1−S(t))S'(t) = S(t)(1 - S(t)), is **true**!

STEP 13

The fourth equation, S′(t)=−S(−t)S'(t) = -S(-t), is not going to work out either, since we'd have a negative on one side and a positive on the other.

STEP 14

S′(t)=e−t(1+e−t)2S'(t) = \frac{e^{-t}}{(1 + e^{-t})^{2}}.
The correct equation is S′(t)=S(t)(1−S(t))S'(t) = S(t)(1 - S(t)).

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