How to find an equation of the tangent line to the curve at the given point if #y=cosx-sinx# and #(pi,-1)#?

Question

Evelyn Agard · Accepted Answer

First evaluate the derivative of your function:
#y'(x)=-sin(x)-cos(x)#
Then evaluate the derivative at your point, i.e., at #x=pi# to find the slope #m# of the tangent line:
#m=y'(pi)=0+1=1#
Finally, use the relationship:
#y-y_0=m(x-x_0)# to find the equation of the tangent line:
#y-(-1)=1(x-pi)#
#y=x-(pi+1)#

Graphically:

Adam Adkins · Answer

undefined To find the equation of the tangent line to the curve y = cos(x) - sin(x) at the point (π, -1), we can use the derivative of the function.

First, find the derivative of y with respect to x, which is dy/dx = -sin(x) - cos(x).

Next, substitute the x-coordinate of the given point (π) into the derivative to find the slope of the tangent line at that point.

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

So, the slope of the tangent line is 1.

Now, we have the slope (m = 1) and the point (π, -1). We can use the point-slope form of a linear equation to find the equation of the tangent line.

y - y1 = m(x - x1), where (x1, y1) is the given point.

Plugging in the values, we get:

y - (-1) = 1(x - π).

Simplifying, we have:

y + 1 = x - π.

Rearranging the equation, we get the equation of the tangent line:

y = x - π - 1.