Math Snap

STEP 1

What is this asking?
We need to find the derivative of the function $f(x) = 3^{-x}$, which means we're looking for the instantaneous rate of change of $f(x)$ at any given $x$.
Watch out!
Negative exponents can be tricky!
Don't forget the chain rule, and remember your exponent rules.

STEP 2

1. Rewrite with e
2. Apply the chain rule
3. Simplify

STEP 3

Let's rewrite our function $f(x) = 3^{-x}$ using the magical number e!
Remember, we can rewrite any exponential function $a^x$ as $e^{\ln(a) \cdot x}$.
Why? Because $e^{\ln(a)} = a$!
It's like a secret code!

STEP 4

So, for our function, we have $f(x) = 3^{-x} = e^{\ln(3) \cdot (-x)} = e^{-\ln(3) \cdot x}$.
See how we brought that negative sign out front? Clean and crisp!

STEP 5

Now, it's chain rule time!
Remember, the chain rule says that the derivative of a composite function like $e^{u(x)}$ is $e^{u(x)} \cdot u'(x)$.
It's like peeling an onion, layer by layer!

STEP 6

In our case, $u(x) = -\ln(3) \cdot x$, so $u'(x) = -\ln(3)$.
See? The derivative of $x$ is just 1, and the $-\ln(3)$ is just a constant hanging out.

STEP 7

Putting it all together, we get $f'(x) = e^{-\ln(3) \cdot x} \cdot (-\ln(3))$. Boom!

STEP 8

Let's make this look nice and tidy.
Remember how we rewrote $3^{-x}$ as $e^{-\ln(3) \cdot x}$?
Well, now we can switch it back!

STEP 9

So, we have $f'(x) = 3^{-x} \cdot (-\ln(3))$.
We can write this even more neatly as $f'(x) = -\ln(3) \cdot 3^{-x}$. Perfect!

The derivative of $f(x) = 3^{-x}$ is $f'(x) = -\ln(3) \cdot 3^{-x}$.