Find `dy/dx if y= sinx+ tany`

Answer

To find \(\frac{dy}{dx}\) for the equation \(y = \sin x + \tan y\), we will differentiate both sides with respect to \(x\).

### Step-by-Step Solution:

1. **Differentiate both sides**:
We start with the equation:
\[ y = \sin x + \tan y \]
Now, we differentiate both sides with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sin x) + \frac{d}{dx}(\tan y) \]

2. **Apply the differentiation rules**:
The derivative of \(\sin x\) is \(\cos x\). For \(\tan y\), we use the chain rule:
\[ \frac{d}{dx}(\tan y) = \sec^2 y \cdot \frac{dy}{dx} \]
Therefore, substituting these derivatives into our equation gives:
\[ \frac{dy}{dx} = \cos x + \sec^2 y \cdot \frac{dy}{dx} \]

3. **Rearranging the equation**:
Now, we want to isolate \(\frac{dy}{dx}\). We can rearrange the equation:
\[ \frac{dy}{dx} - \sec^2 y \cdot \frac{dy}{dx} = \cos x \]
Factoring out \(\frac{dy}{dx}\) from the left side:
\[ \frac{dy}{dx}(1 - \sec^2 y) = \cos x \]

4. **Solve for \(\frac{dy}{dx}\)**:
Now, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\cos x}{1 - \sec^2 y} \]

5. **Final expression**:
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos x}{1 - \sec^2 y} \]