use implicit differentiation to find dy/dx. 7 cos(xy)=6x + 5y. dy/dx =

Explanation:

Step1: Differentiate both sides

Differentiate $7\cos(xy)$ and $6x + 5y$ with respect to $x$. For the left - hand side, use the chain rule and product rule. The derivative of $\cos(u)$ with respect to $x$ is $-\sin(u)\cdot u'$, where $u = xy$. By the product rule, if $u=xy$, then $u'=y + x\frac{dy}{dx}$. So the derivative of $7\cos(xy)$ is $-7\sin(xy)\left(y + x\frac{dy}{dx}\right)$. The derivative of the right - hand side: the derivative of $6x$ with respect to $x$ is 6, and the derivative of $5y$ with respect to $x$ is $5\frac{dy}{dx}$. So we have: $$-7\sin(xy)\left(y + x\frac{dy}{dx}\right)=6 + 5\frac{dy}{dx}$$

Step2: Expand the left - hand side

Expand $-7\sin(xy)\left(y + x\frac{dy}{dx}\right)$: $$-7y\sin(xy)-7x\sin(xy)\frac{dy}{dx}=6 + 5\frac{dy}{dx}$$

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

Move all terms with $\frac{dy}{dx}$ to one side: $$-7x\sin(xy)\frac{dy}{dx}-5\frac{dy}{dx}=6 + 7y\sin(xy)$$ Factor out $\frac{dy}{dx}$ on the left - hand side: $$\frac{dy}{dx}(-7x\sin(xy)-5)=6 + 7y\sin(xy)$$

Step4: Solve for $\frac{dy}{dx}$

Divide both sides by $-7x\sin(xy)-5$: $$\frac{dy}{dx}=\frac{6 + 7y\sin(xy)}{-7x\sin(xy)-5}=-\frac{6 + 7y\sin(xy)}{7x\sin(xy)+5}$$

Answer:

$-\frac{6 + 7y\sin(xy)}{7x\sin(xy)+5}$