Question Statement

Find $\frac{dy}{dx}$ given that $x+y=\cos (xy)$.

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Background and Explanation

These problems require 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

Q14: Differentiating $x+y=\cos (xy)$

We begin with the equation: $x+y=\cos (xy)$

Differentiate both sides with respect to $x$:

$\frac{d}{dx}(x)+\frac{d}{dx}(y)=\frac{d}{dx}[\cos (xy)]$

For the left side, we obtain: $1+\frac{dy}{dx}$

For the right side, apply the chain rule. The derivative of $\cos (u)$ is $-\sin (u)\cdot \frac{du}{dx}$ where $u=xy$: $-\sin (xy)\cdot \frac{d}{dx}(xy)$

Apply the product rule to differentiate $xy$: $\frac{d}{dx}(xy)=y\cdot \frac{d}{dx}(x)+x\cdot \frac{d}{dx}(y)=y(1)+x\frac{dy}{dx}=y+x\frac{dy}{dx}$

Substituting this back: $1+\frac{dy}{dx}=-\sin (xy)\cdot \left(y+x\frac{dy}{dx}\right)$

Expand the right side by distributing $-\sin (xy)$: $1+\frac{dy}{dx}=-y\sin (xy)-x\sin (xy)\frac{dy}{dx}$

Collect all terms containing $\frac{dy}{dx}$ on the left side and the remaining terms on the right: $\frac{dy}{dx}+x\sin (xy)\frac{dy}{dx}=-y\sin (xy)-1$

Factor out $\frac{dy}{dx}$ from the left side: $(1+x\sin (xy))\frac{dy}{dx}=-(y\sin (xy)+1)$

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

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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: $\frac{d}{dx}[f(y)]=f^{\prime }(y)\cdot \frac{dy}{dx}$ and $\frac{d}{dx}[f(xy)]=f^{\prime }(xy)\cdot \frac{d}{dx}(xy)$
• Product Rule: $\frac{d}{dx}[uv]=u\frac{dv}{dx}+v\frac{du}{dx}$ (applied to products like $xy$, $x\sin y$, and $y\cos x$)
• Trigonometric Derivatives: $\frac{d}{dx}(\sin u)=\cos u\cdot \frac{du}{dx}$ and $\frac{d}{dx}(\cos u)=-\sin u\cdot \frac{du}{dx}$

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Summary of Steps

1. Differentiate both sides with respect to $x$, treating $y$ as an implicit function of $x$.
2. Apply the chain rule whenever differentiating a term containing $y$ (multiply by $\frac{dy}{dx}$).
3. Apply the product rule for any products involving both $x$ and $y$.
4. Expand and simplify the resulting expressions.
5. Collect all $\frac{dy}{dx}$ terms on one side of the equation and all other terms on the opposite side.
6. Factor out $\frac{dy}{dx}$ and solve algebraically to isolate the derivative.