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How do you find the equation of the line tangent to #y=cosx-sinx#, at (pi,-1)?

Answer 1

Elijah Abram

#y=x-(pi+1)#

#•color(white)(2/2)dy/dx=m_(color(red)"tangent")" at x = a"#

#y=cosx-sinx#

#rArrdy/dx=-sinx-cosx#

#x=pitody/dx=-sinpi-cospi=1#

#rArry+1=x-pi#

#rArry=x-(pi+1)larr" equation of tangent"#

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Answer 2

Aurora Alvarado

To find the equation of the line tangent to the curve y = cos(x) - sin(x) at the point (π, -1), we need to determine the slope of the tangent line at that point.

First, we find the derivative of the given function y = cos(x) - sin(x) with respect to x.

dy/dx = -sin(x) - cos(x)

Next, we substitute π into the derivative to find the slope at x = π.

dy/dx = -sin(π) - cos(π) = 0 - (-1) = 1

Therefore, the slope of the tangent line at (π, -1) is 1.

Using the point-slope form of a linear equation, we can write the equation of the tangent line as:

y - y1 = m(x - x1)

Substituting the values of (π, -1) and m = 1, we have:

y - (-1) = 1(x - π)

Simplifying, we get:

y + 1 = x - π

Rearranging the equation, we have:

y = x - π - 1

Thus, the equation of the line tangent to y = cos(x) - sin(x) at (π, -1) is y = x - π - 1.

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Answer from HIX Tutor

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.