Find `(dy)/(dx)`, when: `y=(3x+5)^((2x-3))`

Answer

To find \(\frac{dy}{dx}\) for the function \(y = (3x + 5)^{(2x - 3)}\), we will use logarithmic differentiation. Here are the steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides:
\[ \ln y = \ln((3x + 5)^{(2x - 3)}) \]

### Step 2: Apply the logarithmic power rule
Using the property of logarithms that states \(\ln(a^b) = b \ln a\), we can simplify the equation:
\[ \ln y = (2x - 3) \ln(3x + 5) \]

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate both sides. Remember to use the product rule on the right side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}((2x - 3) \ln(3x + 5)) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}((2x - 3) \ln(3x + 5)) \]

### Step 4: Apply the product rule on the right side
Using the product rule:
\[ \frac{d}{dx}((2x - 3) \ln(3x + 5)) = (2x - 3) \frac{d}{dx}(\ln(3x + 5)) + \ln(3x + 5) \frac{d}{dx}(2x - 3) \]
Calculating the derivatives:
\[ \frac{d}{dx}(\ln(3x + 5)) = \frac{3}{3x + 5} \quad \text{and} \quad \frac{d}{dx}(2x - 3) = 2 \]
Substituting these into our equation:
\[ \frac{1}{y} \frac{dy}{dx} = (2x - 3) \left(\frac{3}{3x + 5}\right) + \ln(3x + 5)(2) \]

### Step 5: Multiply through by \(y\)
Now, we multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y \left((2x - 3) \frac{3}{3x + 5} + 2 \ln(3x + 5)\right) \]

### Step 6: Substitute back for \(y\)
Since \(y = (3x + 5)^{(2x - 3)}\), we substitute back:
\[ \frac{dy}{dx} = (3x + 5)^{(2x - 3)} \left((2x - 3) \frac{3}{3x + 5} + 2 \ln(3x + 5)\right) \]

### Final Answer
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = (3x + 5)^{(2x - 3)} \left((2x - 3) \frac{3}{3x + 5} + 2 \ln(3x + 5)\right) \]