If `x^2+x y+y^2=7/4,` then `(dy)/(dx)` at `x=1` and `y=1/2` is:

Options

A. `3/4`
B. `(-5)/4`
C. `21/8`
D. `(-21)/8`

Answer

To find \(\frac{dy}{dx}\) at the point \(x = 1\) and \(y = \frac{1}{2}\) for the equation \(x^2 + xy + y^2 = \frac{7}{4}\), we will follow these steps:

### Step 1: Differentiate the equation implicitly
We start with the equation:
\[ x^2 + xy + y^2 = \frac{7}{4} \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}\left(\frac{7}{4}\right) \]

### Step 2: Apply differentiation rules
The differentiation of each term gives:
- For \(x^2\), the derivative is \(2x\).
- For \(xy\), we use the product rule: \(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\).
- For \(y^2\), we use the chain rule: \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\).
- The derivative of a constant \(\frac{7}{4}\) is \(0\).

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

### Step 3: Rearranging the equation
Now, we can rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ 2x + y + \left(x + 2y\right)\frac{dy}{dx} = 0 \]
\[ \left(x + 2y\right)\frac{dy}{dx} = - (2x + y) \]
\[ \frac{dy}{dx} = -\frac{2x + y}{x + 2y} \]

### Step 4: Substitute the values of \(x\) and \(y\)
Now we substitute \(x = 1\) and \(y = \frac{1}{2}\) into the expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = -\frac{2(1) + \frac{1}{2}}{1 + 2\left(\frac{1}{2}\right)} \]

### Step 5: Simplifying the expression
Calculating the numerator:
\[ 2(1) + \frac{1}{2} = 2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2} \]
Calculating the denominator:
\[ 1 + 2\left(\frac{1}{2}\right) = 1 + 1 = 2 \]
So, we have:
\[ \frac{dy}{dx} = -\frac{\frac{5}{2}}{2} = -\frac{5}{4} \]

### Final Answer
Thus, the value of \(\frac{dy}{dx}\) at \(x = 1\) and \(y = \frac{1}{2}\) is:
\[ \frac{dy}{dx} = -\frac{5}{4} \]