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## Homework Statement

Determine the slope of the tangent at x = 0 for the function f(x) = [tex]\frac{cosx}{1-x}[/tex]?

## Homework Equations

Product rule:

F'(x) = f'(x)g(x)+f(x)g'(x)

Chain rule:

f'(x) = nx^n-1·(x)'

## The Attempt at a Solution

So first I rewrite the equation to get rid of the fraction:

f(x) = (cosx)(x-1)^-1

Then I start differentiating:

f'(x) = (cosx)'(x-1)^-1 + (cosx)(1-x)^-1'

f'(x) = (-sinx)(x-1)^-1 + (cosx)(-1)(1-x)^(-1-1)·(-1)

f'(x) = -sin/(1-x) + (cosx)/(1-x)^2

f'(x) = -sinx(1-x)+(cosx)/(1-x)^2

That is what I calculate as the derivative.

So when I substitute in 0:

f(0) = -sin0(1-0)+(cos0)/(1-0)^2 = 0/1 = 0. The problem is, my book says that the slope is 1, which would mean I should get 1/1, but how? Did I miscalculate during the differentiation? Unfortunately, I am not allowed to use quotient rule according to my teacher to solve these types of problems. Any help? Thanks in advance.