# determine slope of tangent

• September 1st 2010, 09:33 AM
euclid2
determine slope of tangent
determine the slope of the tangent at x=0 for the function $f(x)=\frac{cos x}{1 - x}$

i got the derivative to be

$f'(x)=\frac{((-sinx)(1-x)-(cosx)(-1))}{(1-x)^2}$

$f'(x)=\frac{((-sinx)(1-x)+cosx))}{(1-x)^2}$
• September 1st 2010, 09:48 AM
Quote:

Originally Posted by euclid2
determine the slope of the tangent at x=0 for the function $f(x)=\frac{cos x}{1 - x}$

i got the derivative to be

$f'(x)=\frac{((-sinx)(1-x)-(cosx)(-1))}{(1-x)^2}$

$f'(x)=\frac{((-sinx)(1-x)+cosx))}{(1-x)^2}$

Placing x=0 into the derivative, you have $f'(x)=\displaystyle\frac{1}{1}$

therefore the tangent at x=0 is the line y=x+1