Question Statement

Find $\frac{dy}{dx}$ given that: $x\sin y-y\cos x=1$

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Background and Explanation

This problem requires implicit differentiation, a technique used when $y$ is defined implicitly as a function of $x$ rather than explicitly solved for $y$. We differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$ and applying the chain rule whenever we differentiate terms containing $y$.

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Solution

Differentiating $x\sin y-y\cos x=1$

We start with the given equation: $x\sin y-y\cos x=1$

Differentiate both sides with respect to $x$: $\frac{d}{dx}(x\sin y)-\frac{d}{dx}(y\cos x)=\frac{d}{dx}(1)$

Step 1: Apply the product rule to the first term $x\sin y$

Using the rule $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$, where $u=x$ and $v=\sin y$: $\frac{d}{dx}(x)\cdot \sin y+x\cdot \frac{d}{dx}(\sin y)$ Since $y$ is a function of $x$, we use the chain rule for $\frac{d}{dx}(\sin y)$: $1\cdot \sin y+x\cdot \cos y\cdot \frac{dy}{dx}=\sin y+x\cos y\frac{dy}{dx}$

Step 2: Apply the product rule to the second term $y\cos x$

Using the rule where $u=y$ and $v=\cos x$: $\frac{d}{dx}(y)\cdot \cos x+y\cdot \frac{d}{dx}(\cos x)$ $\frac{dy}{dx}\cdot \cos x+y\cdot (-\sin x)=\frac{dy}{dx}\cos x-y\sin x$

Step 3: Combine the derivatives

Substitute these results back into the differentiated equation. Be careful to distribute the negative sign to both parts of the second term: $\left(\sin y+x\cos y\frac{dy}{dx}\right)-\left(\frac{dy}{dx}\cos x-y\sin x\right)=0$

Remove the parentheses: $\sin y+x\cos y\frac{dy}{dx}-\frac{dy}{dx}\cos x+y\sin x=0$

Step 4: Isolate $\frac{dy}{dx}$

Group the terms containing $\frac{dy}{dx}$ on the left side and move all other terms to the right side: $x\cos y\frac{dy}{dx}-\cos x\frac{dy}{dx}=-y\sin x-\sin y$

Factor out $\frac{dy}{dx}$ from the left side: $(x\cos y-\cos x)\frac{dy}{dx}=-(y\sin x+\sin y)$

Finally, solve for $\frac{dy}{dx}$ by dividing both sides by $(x\cos y-\cos x)$: $\frac{dy}{dx}=\frac{-(y\sin x+\sin y)}{x\cos y-\cos x}$

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Key Formulas or Methods Used

• Implicit Differentiation: Differentiating both sides of an equation with respect to $x$ while treating $y$ as a function of $x$.
• Chain Rule: $\frac{d}{dx}[f(y)]=f^{\prime }(y)\cdot \frac{dy}{dx}$.
• Product Rule: $\frac{d}{dx}[uv]=u\frac{dv}{dx}+v\frac{du}{dx}$.
• Trigonometric Derivatives:
  • $\frac{d}{dx}(\sin u)=\cos u\cdot \frac{du}{dx}$
  • $\frac{d}{dx}(\cos u)=-\sin u\cdot \frac{du}{dx}$

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Summary of Steps

1. Differentiate both sides of the equation with respect to $x$.
2. Apply the product rule to the terms $x\sin y$ and $y\cos x$.
3. Apply the chain rule whenever differentiating $y$, resulting in a $\frac{dy}{dx}$ term.
4. Distribute negative signs carefully when simplifying the expression.
5. Collect all terms with $\frac{dy}{dx}$ on one side and move the remaining terms to the other.
6. Factor out $\frac{dy}{dx}$ and divide to isolate the final derivative.