If `2x^(2)y^(2)-3xy+1=0," then "((dy)/(dx))" at "(1,1)` is

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To find \(\frac{dy}{dx}\) at the point \((1, 1)\) for the equation \(2x^2y^2 - 3xy + 1 = 0\), we will use implicit differentiation. Here are the steps:

### Step 1: Differentiate the equation implicitly
We start with the equation:
\[ 2x^2y^2 - 3xy + 1 = 0 \]
Now, we differentiate both sides with respect to \(x\).

Using the product rule for \(2x^2y^2\) and \(-3xy\):
\[ \frac{d}{dx}(2x^2y^2) = 2 \cdot (x^2 \cdot \frac{d}{dx}(y^2) + y^2 \cdot \frac{d}{dx}(x^2)) = 2(2xy^2\frac{dy}{dx} + 2x^2y) \]
\[ \frac{d}{dx}(-3xy) = -3 \left( x\frac{dy}{dx} + y \right) \]
The derivative of the constant \(1\) is \(0\).

Putting it all together, we have:
\[ 2(2xy^2\frac{dy}{dx} + 2x^2y) - 3\left(x\frac{dy}{dx} + y\right) = 0 \]

### Step 2: Simplify the expression
Expanding and simplifying gives:
\[ 4xy^2\frac{dy}{dx} + 4x^2y - 3x\frac{dy}{dx} - 3y = 0 \]

### Step 3: Collect terms involving \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ (4xy^2 - 3x)\frac{dy}{dx} + (4x^2y - 3y) = 0 \]
\[ (4xy^2 - 3x)\frac{dy}{dx} = - (4x^2y - 3y) \]

### Step 4: Solve for \(\frac{dy}{dx}\)
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{-(4x^2y - 3y)}{(4xy^2 - 3x)} \]

### Step 5: Substitute the point \((1, 1)\)
Now we substitute \(x = 1\) and \(y = 1\):
\[ \frac{dy}{dx} = \frac{-(4(1)^2(1) - 3(1))}{(4(1)(1)^2 - 3(1))} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{-(4 - 3)}{(4 - 3)} = \frac{-1}{1} = -1 \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at the point \((1, 1)\) is:
\[ \frac{dy}{dx} = -1 \]