Product Rule

Today, we are going to talk about the Product Rule. Does anyone know what it refers to?

Is it about multiplying two functions together?

Exactly! The Product Rule helps us find the derivative when we multiply two functions. It states that if $f(x) = g(x) \cdot h(x)$, then $f'(x) = g'(x)h(x) + g(x)h'(x)$.

Can you give an example?

Sure! If we have $f(x) = x^2 \cdot ext{sin}(x)$, using the Product Rule we differentiate to get $f'(x) = 2x ext{sin}(x) + x^2 ext{cos}(x)$.

Why do we need both derivatives?

Great question! We need both derivatives because they account for the changes in both functions independently and ensure we capture the complete rate of change.

Can you summarize what we have learned?

Certainly! The Product Rule allows us to differentiate products of functions by combining the derivative of one function with the other function itself. Always remember the format: $f'(x) = g'(x)h(x) + g(x)h'(x)$.

Now that we understand the Product Rule, let's try applying it together. How would we differentiate $f(x) = x^3 ext{cos}(x)$?

First, we identify $g(x) = x^3$ and $h(x) = ext{cos}(x)$.

Correct! What are the derivatives of these functions?

$g'(x) = 3x^2$ and $h'(x) = - ext{sin}(x)$.

Great! Now apply the Product Rule. What do we get?

We get $f'(x) = 3x^2 ext{cos}(x) + x^3 (- ext{sin}(x))$ = $3x^2 ext{cos}(x) - x^3 ext{sin}(x)$.

Excellent job! This example shows how the Product Rule allows us to calculate complex derivatives efficiently.

Let's solidify our understanding with a practice problem. Differentiate $f(x) = e^x ext{ln}(x)$.

We set $g(x) = e^x$ and $h(x) = ext{ln}(x)$.

Perfect! What are the derivatives?

$g'(x) = e^x$ and $h'(x) = \frac{1}{x}$.

Now apply the Product Rule.

So, we have $f'(x) = e^x ext{ln}(x) + e^x \cdot \frac{1}{x}$.

Excellent! You can further simplify it to get your final answer.

The final answer is $f'(x) = e^x ( ext{ln}(x) + \frac{1}{x})$.

Great teamwork, everyone! You've successfully applied the Product Rule!