Question

Differentiate x^2 + y^2-2x-6y+5. Find dy/dx and d^2y/dx^2 at x =3 and y =2

Answer 1

To differentiate the expression \( x^2 + y^2 - 2x - 6y + 5 = 0 \) implicitly with respect to \( x \), we will take the derivative of each term with respect to \( x \) and solve for \( \frac{dy}{dx} \).

1. **Differentiate the equation:**
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(2x) - \frac{d}{dx}(6y) + \frac{d}{dx}(5) = 0
\]
Using the chain rule for \( y^2 \) and \( y \):
\[
2x + 2y \frac{dy}{dx} - 2 - 6 \frac{dy}{dx} = 0
\]

2. **Rearranging the equation:**
\[
2x - 2 + (2y - 6) \frac{dy}{dx} = 0
\]
\[
(2y - 6) \frac{dy}{dx} = 2 - 2x
\]
\[
\frac{dy}{dx} = \frac{2 - 2x}{2y - 6}
\]

3. **Evaluate \( \frac{dy}{dx} \) at \( x = 3 \) and \( y = 2 \):**
Substitute \( x = 3 \) and \( y = 2 \) into the derivative:
\[
\frac{dy}{dx} = \frac{2 - 2(3)}{2(2) - 6} = \frac{2 - 6}{4 - 6} = \frac{-4}{-2} = 2
\]

4. **Now, we need to find the second derivative \( \frac{d^2y}{dx^2} \):**
To find \( \frac{d^2y}{dx^2} \), we differentiate \( \frac{dy}{dx} \) implicitly with respect to \( x \):
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{2 - 2x}{2y - 6}\right)
\]
We will use the quotient rule here. Let \( u = 2 - 2x \) and \( v = 2y - 6 \):
\[
\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]
where
\(\frac{du}{dx} = -2\) and \(\frac{dv}{dx} = 2 \frac{dy}{dx}\).

Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[
v = 2y - 6 = 2(2) - 6 = 4 - 6 = -2,
\]
\[
\frac{dy}{dx} = 2,
\]
\[
\frac{dv}{dx} = 2 \cdot 2 = 4.
\]

Now substitute \( u, v, \frac{du}{dx}, \frac{dv}{dx} \) into the quotient rule:
\[
\frac{d^2y}{dx^2} = \frac{(-2)(-2) - (2 - 6)(4)}{(-2)^2} = \frac{4 - (-4)}{4} = \frac{4 + 4}{4} = \frac{8}{4} = 2.
\]

### Final Answers:
- \( \frac{dy}{dx} = 2 \)
- \( \frac{d^2y}{dx^2} = 2 \) at \( x = 3 \) and \( y = 2 \)