Question Statement

Find $\frac{dy}{dx}$ given the implicit equation:

$xy=\sin x+y$

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Background and Explanation

This problem requires implicit differentiation, where we differentiate both sides of an equation with respect to $x$ while treating $y$ as a function of $x$ (so every $y$ term requires the chain rule). You will need the product rule to differentiate the term $xy$.

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Solution

Start with the given equation:

$xy=\sin x+y$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(xy)=\frac{d}{dx}(\sin x)+\frac{d}{dx}(y)$

Apply the product rule to the left side $\frac{d}{dx}(xy)$. Recall that $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, $u=x$ and $v=y$:

$y\cdot 1+x\cdot \frac{dy}{dx}=\cos x+\frac{dy}{dx}$

Simplify the left side:

$y+x\frac{dy}{dx}=\cos x+\frac{dy}{dx}$

Rearrange the equation to collect all terms containing $\frac{dy}{dx}$ on the left side and remaining terms on the right. Subtract $\frac{dy}{dx}$ from both sides:

$y+x\frac{dy}{dx}-\frac{dy}{dx}=\cos x$

Now subtract $y$ from both sides to isolate the derivative terms:

$x\frac{dy}{dx}-\frac{dy}{dx}=\cos x-y$

Factor out $\frac{dy}{dx}$ from the left side:

$(x-1)\frac{dy}{dx}=\cos x-y$

Finally, solve for $\frac{dy}{dx}$ by dividing both sides by $(x-1)$:

$\frac{dy}{dx}=\frac{\cos x-y}{x-1}$

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Key Formulas or Methods Used

• Implicit differentiation: Differentiating both sides with respect to $x$ while treating $y$ as $y(x)$
• Product rule: $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$ (used for the term $xy$)
• Chain rule: $\frac{d}{dx}(y)=\frac{dy}{dx}$ (applied to all $y$ terms)
• Derivative of sine: $\frac{d}{dx}(\sin x)=\cos x$

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Summary of Steps

1. Differentiate both sides of $xy=\sin x+y$ with respect to $x$.
2. Apply the product rule to $xy$ to obtain $y+x\frac{dy}{dx}$ on the left side.
3. Differentiate the right side to get $\cos x+\frac{dy}{dx}$.
4. Rearrange terms to group all $\frac{dy}{dx}$ terms on one side and remaining terms on the other.
5. Factor out $\frac{dy}{dx}$ and divide to isolate the derivative.