Given that `y= (3x -1)^(2) + (2x -1)^(3)`, find `(dy)/( dx)` and the 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 the function
We start with the function:
\[ y = (3x - 1)^2 + (2x - 1)^3 \]

Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}[(3x - 1)^2] + \frac{d}{dx}[(2x - 1)^3] \]

### Step 2: Apply the chain rule
Using the chain rule for differentiation, we find:
1. For \( (3x - 1)^2 \):
\[ \frac{d}{dx}[(3x - 1)^2] = 2(3x - 1) \cdot \frac{d}{dx}[3x - 1] = 2(3x - 1) \cdot 3 = 6(3x - 1) \]

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

### Step 3: Combine the derivatives
Now, we combine the results from the two derivatives:
\[ \frac{dy}{dx} = 6(3x - 1) + 6(2x - 1)^2 \]

### Step 4: Factor out common terms
We can factor out the common term \( 6 \):
\[ \frac{dy}{dx} = 6 \left( (3x - 1) + (2x - 1)^2 \right) \]

### Step 5: Set the derivative to zero
To find the points where \( \frac{dy}{dx} = 0 \):
\[ 6 \left( (3x - 1) + (2x - 1)^2 \right) = 0 \]
This simplifies to:
\[ (3x - 1) + (2x - 1)^2 = 0 \]

### Step 6: Expand and simplify
Expanding \( (2x - 1)^2 \):
\[ (2x - 1)^2 = 4x^2 - 4x + 1 \]
Thus, we have:
\[ 3x - 1 + 4x^2 - 4x + 1 = 0 \]
Combining like terms:
\[ 4x^2 - x + 0 = 0 \]

### Step 7: Factor the quadratic equation
Factoring out \( x \):
\[ x(4x - 1) = 0 \]
This gives us:
\[ x = 0 \quad \text{or} \quad 4x - 1 = 0 \Rightarrow x = \frac{1}{4} \]

### Step 8: Find the corresponding y-values
Now we 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 \( x = \frac{1}{4} \):
\[ y = \left(3\left(\frac{1}{4}\right) - 1\right)^2 + \left(2\left(\frac{1}{4}\right) - 1\right)^3 \]
\[ = \left(\frac{3}{4} - 1\right)^2 + \left(\frac{1}{2} - 1\right)^3 = \left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{2}\right)^3 \]
\[ = \frac{1}{16} - \frac{1}{8} = \frac{1}{16} - \frac{2}{16} = -\frac{1}{16} \]
So, the point is \( \left(\frac{1}{4}, -\frac{1}{16}\right) \).

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