use implicit differentiation to find an equation of the tangent line to the curve sin(x + y)=2x - 2y at the…

Explanation:

Step1: Differentiate both sides

Differentiate $\sin(x + y)$ with respect to $x$ using the chain - rule. The derivative of $\sin(u)$ with respect to $x$ is $\cos(u)\cdot u'$, where $u=x + y$ and $u'=1 + y'$. The derivative of $2x-2y$ with respect to $x$ is $2-2y'$. So, $\cos(x + y)\cdot(1 + y')=2-2y'$.

Step2: Expand the left - hand side

Expand $\cos(x + y)\cdot(1 + y')$ to get $\cos(x + y)+\cos(x + y)y'=2-2y'$.

Step3: Isolate $y'$ terms

Move all terms with $y'$ to one side: $\cos(x + y)y'+2y'=2-\cos(x + y)$.

Step4: Factor out $y'$

Factor out $y'$ on the left - hand side: $y'(\cos(x + y)+2)=2-\cos(x + y)$.

Step5: Solve for $y'$

$y'=\frac{2-\cos(x + y)}{\cos(x + y)+2}$.

Step6: Find the slope at the point $(\pi,\pi)$

Substitute $x = \pi$ and $y=\pi$ into $y'$. $\cos(\pi+\pi)=\cos(2\pi)=1$. So, $y'|_{x = \pi,y=\pi}=\frac{2 - 1}{1+2}=\frac{1}{3}$.

Step7: Use the point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(\pi,\pi)$ and $m=\frac{1}{3}$. So, $y-\pi=\frac{1}{3}(x - \pi)$.

Step8: Rearrange to slope - intercept form

$y=\frac{1}{3}x-\frac{\pi}{3}+\pi=\frac{1}{3}x+\frac{2\pi}{3}$.

Answer:

$y=\frac{1}{3}x+\frac{2\pi}{3}$