# How do you find the equation of the tangent line to the curve at the given point #sin(x+y)=2x-2y# at point (pi, pi)?

The equation is

Use a combination of implicit differentiation and the product rule:

We can now use point-slope form to determine the equation of the tangent.

Hopefully this helps!

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To find the equation of the tangent line to the curve at the given point (pi, pi), we need to find the derivative of the curve with respect to x. Taking the derivative of sin(x+y)=2x-2y with respect to x, we get cos(x+y) * (1+dy/dx) = 2.

Next, we substitute the coordinates of the given point (pi, pi) into the equation. Plugging in x=pi and y=pi, we have cos(pi+pi) * (1+dy/dx) = 2. Simplifying this equation gives cos(2pi) * (1+dy/dx) = 2. Since cos(2pi) = 1, the equation becomes 1 * (1+dy/dx) = 2.

Solving for dy/dx, we have 1 + dy/dx = 2. Subtracting 1 from both sides gives dy/dx = 1.

Therefore, the slope of the tangent line at the point (pi, pi) is 1. Using the point-slope form of a line, we can write the equation of the tangent line as y - pi = 1(x - pi), which simplifies to y = x.

Hence, the equation of the tangent line to the curve sin(x+y)=2x-2y at the point (pi, pi) is y = x.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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