Given that `y=(3x-1)^(2)+(2x-1)^(3)`, find `(dy)/(dx)` and points on the curve for which `(dy)/(dx)=0`.

Answer

To solve the problem, we need to find the derivative of the function \( y = (3x - 1)^2 + (2x - 1)^3 \) and then determine the points on the curve where this derivative equals zero.

### Step 1: Differentiate \( y \) with respect to \( x \)

We start with the function:
\[ y = (3x - 1)^2 + (2x - 1)^3 \]

To find \( \frac{dy}{dx} \), we will use the chain rule and the power rule.

1. Differentiate \( (3x - 1)^2 \):
\[ \frac{d}{dx}[(3x - 1)^2] = 2(3x - 1) \cdot \frac{d}{dx}(3x - 1) = 2(3x - 1) \cdot 3 = 6(3x - 1) \]

2. Differentiate \( (2x - 1)^3 \):
\[ \frac{d}{dx}[(2x - 1)^3] = 3(2x - 1)^2 \cdot \frac{d}{dx}(2x - 1) = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2 \]

Now, we can combine these results:
\[ \frac{dy}{dx} = 6(3x - 1) + 6(2x - 1)^2 \]

### Step 2: Factor out common terms

We can factor out the common factor of 6:
\[ \frac{dy}{dx} = 6 \left( (3x - 1) + (2x - 1)^2 \right) \]

### Step 3: Set \( \frac{dy}{dx} = 0 \)

To find the points where the derivative is zero, we set:
\[ 6 \left( (3x - 1) + (2x - 1)^2 \right) = 0 \]

This simplifies to:
\[ (3x - 1) + (2x - 1)^2 = 0 \]

### Step 4: Solve for \( x \)

Now we need to solve the equation:
\[ (3x - 1) + (2x - 1)^2 = 0 \]

First, expand \( (2x - 1)^2 \):
\[ (2x - 1)^2 = 4x^2 - 4x + 1 \]

Substituting this back into the equation gives:
\[ 3x - 1 + 4x^2 - 4x + 1 = 0 \]
\[ 4x^2 - x = 0 \]

Factoring out \( x \):
\[ x(4x - 1) = 0 \]

This gives us two solutions:
1. \( x = 0 \)
2. \( 4x - 1 = 0 \) which leads to \( x = \frac{1}{4} \)

### Step 5: Find corresponding \( y \) values

Now we need to find the corresponding \( y \) values for \( x = 0 \) and \( x = \frac{1}{4} \).

1. For \( x = 0 \):
\[ y = (3(0) - 1)^2 + (2(0) - 1)^3 = (-1)^2 + (-1)^3 = 1 - 1 = 0 \]
So, the point is \( (0, 0) \).

2. For \( x = \frac{1}{4} \):
\[ y = \left(3 \cdot \frac{1}{4} - 1\right)^2 + \left(2 \cdot \frac{1}{4} - 1\right)^3 \]
\[ = \left(\frac{3}{4} - 1\right)^2 + \left(\frac{1}{2} - 1\right)^3 \]
\[ = \left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{2}\right)^3 = \frac{1}{16} - \frac{1}{8} = \frac{1}{16} - \frac{2}{16} = -\frac{1}{16} \]
So, the point is \( \left(\frac{1}{4}, -\frac{1}{16}\right) \).

### Final Answer

The points on the curve for which \( \frac{dy}{dx} = 0 \) are:
1. \( (0, 0) \)
2. \( \left(\frac{1}{4}, -\frac{1}{16}\right) \)