find $\frac{dy}{dx}$ by implicit differentiation. $cos(xy)=sin(x + y)$ $\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate $\cos(xy)$ and $\sin(x + y)$ with respect to $x$. For $\cos(xy)$, by the chain - rule and product - rule, we have $-\sin(xy)\left(y + x\frac{dy}{dx}\right)$. For $\sin(x + y)$, by the chain - rule, we have $\cos(x + y)\left(1+\frac{dy}{dx}\right)$. So, $-\sin(xy)\left(y + x\frac{dy}{dx}\right)=\cos(x + y)\left(1+\frac{dy}{dx}\right)$.

Step2: Expand the equation

Expand the left - hand side: $-y\sin(xy)-x\sin(xy)\frac{dy}{dx}=\cos(x + y)+\cos(x + y)\frac{dy}{dx}$.

Step3: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side: $-x\sin(xy)\frac{dy}{dx}-\cos(x + y)\frac{dy}{dx}=\cos(x + y)+y\sin(xy)$.

Step4: Factor out $\frac{dy}{dx}$

Factor out $\frac{dy}{dx}$ on the left - hand side: $\frac{dy}{dx}\left[-x\sin(xy)-\cos(x + y)\right]=\cos(x + y)+y\sin(xy)$.

Step5: Solve for $\frac{dy}{dx}$

Divide both sides by $-x\sin(xy)-\cos(x + y)$ to get $\frac{dy}{dx}=-\frac{\cos(x + y)+y\sin(xy)}{x\sin(xy)+\cos(x + y)}$.

Answer:

$-\frac{\cos(x + y)+y\sin(xy)}{x\sin(xy)+\cos(x + y)}$