If any=sin(x+y) then prove that y[1+dy/dx]²

Given that \( a = \sin(x + y) \), we need to prove the relation:

\[ y[1 + \left(\frac{dy}{dx}\right)]^2 \]

### Step 1: Implicit differentiation of the given equation

The equation is:

\[ a = \sin(x + y) \]

We will implicitly differentiate both sides with respect to \( x \). Using the chain rule:

\[
\frac{d}{dx}[a] = \frac{d}{dx}[\sin(x + y)]
\]

Since \( a \) is a constant with respect to \( x \), the left-hand side becomes zero:

\[
0 = \cos(x + y) \cdot \frac{d}{dx}(x + y)
\]

Now differentiate \( x + y \):

\[
0 = \cos(x + y) \cdot \left( 1 + \frac{dy}{dx} \right)
\]

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

To solve for \( \frac{dy}{dx} \), divide both sides by \( \cos(x + y) \) (assuming \( \cos(x + y) \neq 0 \)):

\[
1 + \frac{dy}{dx} = 0
\]

This simplifies to:

\[
\frac{dy}{dx} = -1
\]

### Step 3: Prove the given relation

Now, we use the result \( \frac{dy}{dx} = -1 \) in the expression we need to prove:

\[
y\left[ 1 + \left(\frac{dy}{dx}\right)\right]^2 = y \left[ 1 + (-1) \right]^2 = y \cdot (0)^2 = 0
\]

Thus, the left-hand side simplifies to zero. Since we are not provided with a full form of the relation to prove, it seems we have simplified to a condition where \( \frac{dy}{dx} = -1 \), leading to zero. Therefore, further clarification of the problem is needed.

Would you like more details or clarification on the proof?

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Here are 5 related questions:
1. What is implicit differentiation, and how does it work?
2. How do we apply the chain rule in implicit differentiation?
3. Why do we assume \( \cos(x + y) \neq 0 \) when dividing?
4. What is the significance of solving \( \frac{dy}{dx} = -1 \) in differential equations?
5. How does the derivative change if the equation involves more complex functions of \( y \)?

**Tip**: Always verify that the derivatives satisfy the original equation after solving.

If a = sin(x + y), then prove that y[1 + (dy/dx)]²

Explore the proof of the derivative of a = sin(x + y) using implicit differentiation. This problem includes step-by-step guidance on using the chain rule to find dy/dx and analyze trigonometric functions, suitable for undergraduate calculus students.

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