Questions: To find y' for the function y=(3-x) ln (x+3),

Transcript text: To find $y^{\prime}$ for the function \[ y=(3-x) \ln (x+3), \]

Solution

Solution Steps

To find the derivative $ y' $ of the function $ y = (3 - x) \ln(x + 3) $, we will use the product rule of differentiation. The product rule states that if you have a function $ y = u(x) \cdot v(x) $, then its derivative is $ y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) $. Here, $ u(x) = 3 - x $ and $ v(x) = \ln(x + 3) $.

Solution Approach

Identify $ u(x) $ and $ v(x) $.
Compute $ u'(x) $ and $ v'(x) $.
Apply the product rule: $ y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) $.

Step 1: Define the Function

We start with the function given by \[ y = (3 - x) \ln(x + 3). \]

Step 2: Differentiate the Function

To find the derivative $ y' $, we apply the product rule. Let \[ u = 3 - x \quad \text{and} \quad v = \ln(x + 3). \] Then, we compute the derivatives: \[ u' = -1 \quad \text{and} \quad v' = \frac{1}{x + 3}. \]

Step 3: Apply the Product Rule

Using the product rule, we have: \[ y' = u'v + uv' = (-1) \ln(x + 3) + (3 - x) \cdot \frac{1}{x + 3}. \] This simplifies to: \[ y' = -\ln(x + 3) + \frac{3 - x}{x + 3}. \]

Step 4: Simplify the Expression

The derivative can be expressed as: \[ y' = \frac{3 - x}{x + 3} - \ln(x + 3). \]