find $\frac{dy}{dx}$ by implicit differentiation. $cos(xy)=sin(x + y)$ $\frac{dy}{dx}=$ 6. -/1 points find…

Explanation:

Step1: Differentiate left - hand side

Differentiate $\cos(xy)$ with respect to $x$ using the chain rule and product rule. The derivative of $\cos(u)$ with respect to $x$ is $-\sin(u)\cdot\frac{du}{dx}$, where $u = xy$. By the product rule, $\frac{d(xy)}{dx}=y + x\frac{dy}{dx}$. So, $\frac{d}{dx}\cos(xy)=-\sin(xy)\left(y + x\frac{dy}{dx}\right)$.

Step2: Differentiate right - hand side

Differentiate $\sin(x + y)$ with respect to $x$ using the chain rule. The derivative of $\sin(u)$ with respect to $x$ is $\cos(u)\cdot\frac{du}{dx}$, where $u=x + y$. So, $\frac{d}{dx}\sin(x + y)=\cos(x + y)\left(1+\frac{dy}{dx}\right)$.

Step3: Set derivatives equal

Set $-\sin(xy)\left(y + x\frac{dy}{dx}\right)=\cos(x + y)\left(1+\frac{dy}{dx}\right)$. Expand: $-y\sin(xy)-x\sin(xy)\frac{dy}{dx}=\cos(x + y)+\cos(x + y)\frac{dy}{dx}$.

Step4: Isolate $\frac{dy}{dx}$ terms

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

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

$\frac{dy}{dx}=\frac{\cos(x + y)+y\sin(xy)}{-x\sin(xy)-\cos(x + y)}=-\frac{\cos(x + y)+y\sin(xy)}{x\sin(xy)+\cos(x + y)}$.

Answer:

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

Explanation for second part:

Step1: Differentiate left - hand side

Differentiate $e^{x/y}$ with respect to $x$ using the chain rule and quotient rule. Let $u=\frac{x}{y}$, then $\frac{d}{dx}e^{x/y}=e^{x/y}\cdot\frac{d}{dx}\left(\frac{x}{y}\right)$. By the quotient rule, $\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{y - x\frac{dy}{dx}}{y^{2}}$. So, $\frac{d}{dx}e^{x/y}=e^{x/y}\cdot\frac{y - x\frac{dy}{dx}}{y^{2}}$.

Step2: Differentiate right - hand side

Differentiate $8x - y$ with respect to $x$. The derivative of $8x$ is 8 and the derivative of $-y$ with respect to $x$ is $-\frac{dy}{dx}$. So, $\frac{d}{dx}(8x - y)=8-\frac{dy}{dx}$.

Step3: Set derivatives equal

$e^{x/y}\cdot\frac{y - x\frac{dy}{dx}}{y^{2}}=8-\frac{dy}{dx}$. Multiply through by $y^{2}$: $e^{x/y}(y - x\frac{dy}{dx})=y^{2}(8-\frac{dy}{dx})$. Expand: $ye^{x/y}-xe^{x/y}\frac{dy}{dx}=8y^{2}-y^{2}\frac{dy}{dx}$.

Step4: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side: $-xe^{x/y}\frac{dy}{dx}+y^{2}\frac{dy}{dx}=8y^{2}-ye^{x/y}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}\left(y^{2}-xe^{x/y}\right)=8y^{2}-ye^{x/y}$.

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

$\frac{dy}{dx}=\frac{8y^{2}-ye^{x/y}}{y^{2}-xe^{x/y}}$.

Answer:

$\frac{8y^{2}-ye^{x/y}}{y^{2}-xe^{x/y}}$