`x^(2)/a^2+y^2/b^2=1` find `(dy)/(dx)`

Answer

To find \(\frac{dy}{dx}\) for the equation

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \]

we will use implicit differentiation. Here are the steps:

### Step 1: Differentiate both sides of the equation
Differentiate the left-hand side with respect to \(x\):

\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) + \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = \frac{d}{dx}(1). \]

### Step 2: Apply the differentiation rules
Using the constant multiple rule and the chain rule, we differentiate each term:

1. For \(\frac{x^2}{a^2}\):
\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) = \frac{2x}{a^2}. \]

2. For \(\frac{y^2}{b^2}\):
\[ \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = \frac{2y}{b^2} \cdot \frac{dy}{dx}. \]

3. The derivative of the constant \(1\) is \(0\).

Putting it all together, we have:

\[ \frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0. \]

### Step 3: Rearrange the equation
Now, isolate \(\frac{dy}{dx}\):

\[ \frac{2y}{b^2} \cdot \frac{dy}{dx} = -\frac{2x}{a^2}. \]

### Step 4: Solve for \(\frac{dy}{dx}\)
To solve for \(\frac{dy}{dx}\), divide both sides by \(\frac{2y}{b^2}\):

\[ \frac{dy}{dx} = -\frac{2x}{a^2} \cdot \frac{b^2}{2y}. \]

### Step 5: Simplify the expression
The \(2\) in the numerator and denominator cancels out:

\[ \frac{dy}{dx} = -\frac{x b^2}{y a^2}. \]

Thus, the final answer is:

\[ \frac{dy}{dx} = -\frac{x b^2}{y a^2}. \]