consider the function 3x + sin(xy) = 3. find the slope of the tangent line at the point (1,0). type anything…

Explanation:

Step1: Differentiate both sides

Differentiate $3x+\sin(xy) = 3$ with respect to $x$ using the sum - rule and chain - rule. The derivative of $3x$ with respect to $x$ is $3$. For $\sin(xy)$, by the chain - rule, let $u = xy$, then $\frac{d}{dx}\sin(xy)=\cos(xy)\cdot\frac{d}{dx}(xy)$. Using the product - rule, $\frac{d}{dx}(xy)=y + x\frac{dy}{dx}$. So the derivative of the left - hand side is $3+\cos(xy)\left(y + x\frac{dy}{dx}\right)$, and the derivative of the right - hand side $3$ with respect to $x$ is $0$. So we have $3+\cos(xy)\left(y + x\frac{dy}{dx}\right)=0$.

Step2: Solve for $\frac{dy}{dx}$

Expand the left - hand side: $3 + y\cos(xy)+x\cos(xy)\frac{dy}{dx}=0$. Isolate the terms with $\frac{dy}{dx}$: $x\cos(xy)\frac{dy}{dx}=-3 - y\cos(xy)$. Then $\frac{dy}{dx}=\frac{-3 - y\cos(xy)}{x\cos(xy)}$.

Step3: Substitute the point $(1,0)$

Substitute $x = 1$ and $y = 0$ into $\frac{dy}{dx}$. $\frac{dy}{dx}\big|_{(1,0)}=\frac{-3-0\times\cos(1\times0)}{1\times\cos(1\times0)}=\frac{-3 - 0}{1\times1}=-3$.

Answer:

$-3$