2.5 Q-3

Question Statement

We are asked to find $\frac{d y}{d x}$ for the following two equations:

1. $y = x \cos y$
2. $x = y \sin y$

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Background and Explanation

To solve these problems, we will use the implicit differentiation technique, as the equations involve both $x$ and $y$. Implicit differentiation allows us to differentiate both sides of the equation with respect to $x$, even when $y$ is a function of $x$. We will differentiate the terms involving $y$ as if $y$ is a function of $x$ (using the chain rule).

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Solution

i. Differentiating $y = x \cos y$

1. Start by differentiating both sides of the equation with respect to $x$. $\frac{d}{d x} (y) = \frac{d}{d x} (x \cos y)$

2. On the left-hand side, differentiate $y$ to get $\frac{d y}{d x}$.

   On the right-hand side, we apply the product rule: $\frac{d}{d x} [x \cos y] = x \frac{d}{d x} (\cos y) + \cos y \frac{d}{d x} (x)$

3. Differentiating $\cos y$ with respect to $x$ involves using the chain rule: $\frac{d}{d x} (\cos y) = -\sin y \cdot \frac{d y}{d x}$

4. Differentiating $x$ with respect to $x$ gives $1$.

   Substituting these into the equation: $\frac{d y}{d x} = x \left[ -\sin y \cdot \frac{d y}{d x} \right] + \cos y$

5. Rearranging the terms: $\frac{d y}{d x} + x \sin y \cdot \frac{d y}{d x} = \cos y$

6. Factor out $\frac{d y}{d x}$: $(1 + x \sin y) \frac{d y}{d x} = \cos y$

7. Solving for $\frac{d y}{d x}$: $\frac{d y}{d x} = \frac{\cos y}{1 + x \sin y}$

Thus, the solution for the first part is: $\boxed{\frac{d y}{d x} = \frac{\cos y}{1 + x \sin y}}$

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ii. Differentiating $x = y \sin y$

1. Start by differentiating both sides of the equation with respect to $x$. $\frac{d}{d x} (x) = \frac{d}{d x} (y \sin y)$

2. The left-hand side is straightforward since the derivative of $x$ with respect to $x$ is $1$.

3. On the right-hand side, apply the product rule to differentiate $y \sin y$: $\frac{d}{d x} [y \sin y] = y \frac{d}{d x} (\sin y) + \sin y \frac{d y}{d x}$

4. Differentiating $\sin y$ with respect to $x$ gives: $\frac{d}{d x} (\sin y) = \cos y \cdot \frac{d y}{d x}$

5. Substituting this back: $1 = y \cos y \cdot \frac{d y}{d x} + \sin y \cdot \frac{d y}{d x}$

6. Factor out $\frac{d y}{d x}$: $1 = (y \cos y + \sin y) \cdot \frac{d y}{d x}$

7. Solving for $\frac{d y}{d x}$: $\frac{d y}{d x} = \frac{1}{y \cos y + \sin y}$

Thus, the solution for the second part is: $\boxed{\frac{d y}{d x} = \frac{1}{y \cos y + \sin y}}$

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{d x} (u \cdot v) = u' v + u v'$

• Chain Rule: $\frac{d}{d x} [f(g(x))] = f'(g(x)) \cdot g'(x)$

• Implicit Differentiation: Differentiating both sides of an equation with respect to $x$, where $y$ is implicitly a function of $x$.

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Summary of Steps

1. For $y = x \cos y$:

   • Apply implicit differentiation.
   • Use the product rule on $x \cos y$.
   • Rearrange and solve for $\frac{d y}{d x}$.
   • Result: $\frac{d y}{d x} = \frac{\cos y}{1 + x \sin y}$.

2. For $x = y \sin y$:

   • Apply implicit differentiation.
   • Use the product rule on $y \sin y$.
   • Rearrange and solve for $\frac{d y}{d x}$.
   • Result: $\frac{d y}{d x} = \frac{1}{y \cos y + \sin y}$.