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Let's work through the problems one by one. I'll start with the first one.

### Problem 10: \( x + \sin y = y^2 \)
We are asked to implicitly differentiate this equation with respect to \(x\) to find \(\frac{dy}{dx}\) and use it for the equation of the tangent line.

#### Step 1: Implicit differentiation
Differentiate both sides of the equation with respect to \(x\).

- \( \frac{d}{dx}(x) = 1 \)
- \( \frac{d}{dx}(\sin y) = \cos(y) \cdot \frac{dy}{dx} \) (using chain rule)
- \( \frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx} \) (chain rule again)

So, after differentiating, we get:
\[
1 + \cos(y) \cdot \frac{dy}{dx} = 2y \cdot \frac{dy}{dx}
\]

#### Step 2: Solve for \(\frac{dy}{dx}\)
Now, isolate \(\frac{dy}{dx}\):

\[
\cos(y) \cdot \frac{dy}{dx} - 2y \cdot \frac{dy}{dx} = -1
\]

Factor out \(\frac{dy}{dx}\):

\[
\frac{dy}{dx} (\cos(y) - 2y) = -1
\]

Thus, 
\[
\frac{dy}{dx} = \frac{-1}{\cos(y) - 2y}
\]

This gives us the slope of the tangent line. You can use this equation to find the slope at any point \((x, y)\) on the curve.

#### Step 3: Equation of the tangent line
Once the slope \(m = \frac{dy}{dx}\) is calculated at a specific point, the equation of the tangent line can be written using the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
where \((x_1, y_1)\) is the point of tangency.

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Would you like to proceed with problem 11, or do you want more details on this one? Let me know if you have questions!

### Related questions:
1. How do we apply implicit differentiation to other trigonometric functions?
2. What would happen if we needed the second derivative for the given equation?
3. How do you determine whether a point is on the curve for implicit functions?
4. How can you visualize the tangent line for complex curves like this one?
5. How does implicit differentiation apply to functions involving both \(x\) and \(y\)?

### Tip:
When performing implicit differentiation, be careful to apply the chain rule correctly whenever a function of \(y\) is differentiated with respect to \(x\).

x + sin(y) = y^2, find the equation of the tangent line to the graph of this equation at a given point.

Learn how to find the equation of a tangent line to the graph of the implicit function x + sin(y) = y^2 using implicit differentiation. This problem demonstrates key concepts like the chain rule and solving for dy/dx, suitable for students in Grades 10-12.

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