Bobby Armstrong wrote:
>
> Hello, i need help with this one pronlem,
>
> I need to know the derivative to this equation and the equation to the
line
> that crosses tan to point (0,1)
>
> f(x) = 1 / (x+1)
>
> The answer i keep getting i know isnt right as you can see or i wouldnt
be
> asking people on here.
The derivative of u^n is nu^{n-1} times the derivative of u, so
f'(x) = -1/(x + 1)^2
and the gradient at (0, 1) has slope
f'(0) = -1.
The eqn of the tangent is therefore
y = (-1)x + c
for some c. Since that line p***** through (0, 1)
1 = (-1)0 + c
c = 1.
So the eqn of the tangent at (0, 1) is
y = 1 - x.
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