# determine the slope of the tangent to y=1/(1-x) at the point (2,-1),without constructing a table. Explain what you are doing.

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y=1/(1-x)

To find the slope of the tangent line, we need to determine the derivative of y.

Let u=1 ==> u'=0

v= 1-x ==> v'=-1

==> y=u/v

Now we need to find the derivative y'

y'= (u'v-uv')/v^2.

= [0(1-x) - 1(-1)]/(1-x)^2

y'= 1/(1-x)^2....(1)

Now substitute with x=2 in equation (1)

==> the slope (m)= 1/(1-2)^2 = 1

Now we have the slope m=1 and the point (2,-1).

The formula for the tangent line is:

y=mx+b

Substitute with (2,-1) and m=1

==> -1=1(2)+b

==> b= -3

So, the tangent line is:

y=x-3

To determine the slope of the tangent to y = 1/(1-x) at the point(2,-1).

Solution:

The slope of the tangent is obtained by the value of the differential coefficient dy/dx at the point (2,-1). For this you have to find d/dx{1/(1-x) and then substitute the x=2 and y=-1.

d/dx{1/(1-x)} = [ - 1/(1-x)^2] d/dx(-x) = [-1/(1-x)^2](-1) = 1/(1-x)^2. Now put x= 2 and y =1. in 1/(1-x)^2 = 1/(1-2)^2 = 1/(-1)^2 = 1, ( the y coordinate is absent. So substitution of y=-1 does not arise at all in this case.). Therefore, the slope of the tangent is 1 or the tangeng line at (2,1) makes an angle of arctan (1) or 45 degree (or pi/4 radians) with X axis.

The equation of the tagent line is at the point (x1,y1) is y-y1 = slope at (x1,y1) times (x-x1). Or

y-y1 = [value of dy/dx at (x1,y1)]*(x-x1). Here, in the present problem , (x1,y1) = (2,-1) and (dy/dx) at (2,-1) = 1.

So the equation of the tangent line: y - -1 = 1*(x-2). Or

x-y -3 = 0 is the tangent line.