#### To determine

**To find:** The equation of the tangent line to the given equation at the point.

#### Answer

The equation of the tangent line to the curve sin(x+y)=2x−2y at (π,π) is y=x3+2π3.

#### Explanation

**Given:**

The curve is sin(x+y)=2x−2y.

The point is (π,π).

**Derivative rules:** Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

Where, *m* is the slope of the tangent line at (x1,y1) and m=dydx|(x1,y1).

**Calculation:**

Consider the equation sin(x+y)=2x−2y.

Differentiate the above equation implicitly with respect to *x*,

sin(x+y)=2x−2y

Differentiation with respect to *x*,

ddx(sin(x+y))=ddx(2x−2y)ddx(sin(x+y))=ddx(2x)−ddx(2y)ddx(sin(x+y))=2ddx(x)−2ddx(y)ddx(sin(x+y))=2−2dydx

Let u=x+y and apply the chain rule,

ddx(sinu)=2−2dydxddu(sinu)dudx=2−2dydxcosududx=2−2dydx

Substitute u=x+y,

cos(x+y)ddx(x+y)=2−2dydxcos(x+y)[ddx(x)+dydx]=2−2dydxcos(x+y)[1+dydx]=2−2dydxcos(x+y)+cos(x+y)dydx=2−2dydx

Combine the terms dydx,

cos(x+y)dydx+2dydx=2−cos(x+y)dydx=2−cos(x+y)cos(x+y)+2

Therefore, the derivative of *y* is dydx=2−cos(x+y)cos(x+y)+2.

The slope of the tangent line at (π,π) is computed as follows,

m=dydx|(x,y)=(π,π)=2−cos(π+π)cos(π+π)+2=2−cos2πcos2π+2

Substitute the value cos2π=1,

m=2−11+2=13

Thus, the slope of the tangent line at (π,π) is m=13.

Substitute (π,π) for (x1,y1) and m=13 in equation (1),

y−π=13(x−π)y−π=x3−π3y=x3−π3+πy=x3+2π3

Therefore, the equation of the tangent line to the equation sin(x+y)=2x−2y at (π,π) is y=x3+2π3.