How do you find an equation of the tangent line to the curve at the given point #y = 2xcosx # and #(pi, 2pi)#?
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To find the equation of the tangent line to the curve y = 2xcosx at the point (pi, 2pi), we need to find the slope of the tangent line and then use the pointslope form of a linear equation.

Find the derivative of the function y = 2xcosx using the product rule: dy/dx = 2cosx  2xsinx

Evaluate the derivative at x = pi: dy/dx = 2cos(pi)  2pi*sin(pi) = 2(1)  2pi(0) = 2

The slope of the tangent line is 2.

Use the pointslope form of a linear equation, y  y₁ = m(x  x₁), where (x₁, y₁) is the given point and m is the slope: y  (2pi) = 2(x  pi)

Simplify the equation: y + 2pi = 2x + 2pi

Rearrange the equation to the standard form: 2x + y = 0
Therefore, the equation of the tangent line to the curve y = 2xcosx at the point (pi, 2pi) is 2x + y = 0.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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