# Calculating the slope of the tangent.

-Dragoon-

## Homework Statement

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

## 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.

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cos(0) = 1

Robert1986

## Homework Statement

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

## 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.

Hmmm, what is Sin(0) and what is Cos(0)?

-Dragoon-
Ah, I totally forgot about that. Stupid me. :rofl:

Thanks for all the help, Sammy and robert. 