91Ó°ÊÓ

When differentiating a function that is the product of two or more functions, you need to apply the product rule. In our exercise, the function is given by:
\[ v = x^{3/2} e^{3 x^{3} + 2 x -1} \]

The product rule states that if you have a function that can be written as \( v = u * w \), then its derivative is \( v' = u' w + u w' \).

For our example, we set:

• \( u = x^{3/2} \)
• \( w = e^{3 x^{3} + 2 x - 1} \)

We first need to differentiate each component separately before combining the results using the product rule.
The derivative of \( u \) is \( u' \), and the derivative of \( w \) is \( w' \), which will be computed using another important rule, the chain rule.

The chain rule is used when differentiating a composite function. To apply this rule, you need to identify the outer function and the inner function.

In our example, the function \( w = e^{3 x^{3} + 2 x -1} \) is a composite function where the inner function is \( 3 x^{3} + 2 x - 1 \) and the outer function is the exponential function \( e^x \).

To differentiate \( w \), we use the chain rule as follows:
\[ w = e^{3 x^{3} + 2 x -1} \rightarrow w' = e^{3 x^{3} + 2 x -1} \frac{d}{dx}(3 x^{3} + 2 x - 1) \]
Next, we compute the derivative of the inner function:
\[ \frac{d}{dx}(3 x^{3} + 2 x - 1) = 9 x^{2} + 2 \]
Combining these, we get:
\[ w' = e^{3 x^{3} + 2 x -1} (9 x^{2} + 2) \]
This result tells us how to differentiate the exponential part of our original function.

Derivatives give us the rate at which a function is changing at any given point. In our example, we need to find the derivative of the function:
\[ v = x \right)^{3/2} e^{3 x \right)^{3} + 2 x -1} \ \]

We employ both the product rule and the chain rule to differentiate this function. Firstly, we need to differentiate \( u = x^{3/2} \) using basic rules:

• \( u = x \ right)^{3/2} \ rightarrow u' = \frac{3}{2} x \right)^{1/2} \)

Secondly, for the exponential part \( w \) , we use the chain rule already described. Once we have \( u' \) and \( w \) , we combine them according to the product rule:
\[ v' = u' w + u w' \]

• \( u' = \frac{3}{2} x \ right)^{1/2} \)
• \( w = e^{3 x \ right)^{3} + 2 x - 1} \)
• \( w' = e^{3 x \ right)^{3} + 2 x -1} (9 x^{2} +2)\)
• \( u = x^{3/2}\)

By substituting back, we derive:
\[ v' = \frac{3/2} x \ right)^{1/2} e^{3 x} \ right) ^{3} + 2 x -1} + x \ right)^{3/2} e^{3 x \ right)^{3} + 2 x - 1} (9 x^{2} + 2) \]
Finally, simplify combining similar terms:
\[ v' = e^{3 x \ right) ^{3} +2 x -1 } ( \frac{3}{2} x \ right) ^{1/2} + 9 x \ right) ^{7/2} + 2x \ right) ^{3/2}) \]