Consider this:
$$ K(v) = \frac{v}{v^2+9} $$
Approximate the function for v = 1 by a tangent. I first did the derivitave of the function.
$$ K'(v) = \frac{-v^2+9}{(v^2+9)^2} $$ And now the tangent formula
$$ t(v) = f'(v) * (v - v0) + f(v) $$ Now if I input 1 in both K and K' I get this
$$K'(v) = \frac{-1^2+9}{(1^2+9)^2} $$ That should be 1/10 for K we also get 1/10. Now if I put that in the tangent formula
$$ t(v) = \frac{1}{10} * (v - 1) + \frac{1}{10} $$ Now when I multiply this I only have 1/10v left since 1/10 - 1/10 is 0. Is it possible to have a tangent looking like this,or did I make an error somwhere along the lines.
Thank you!
hint
near $ v_0=1 $, we have the approximation
$$K(1+\epsilon)\approx t(1+\epsilon)$$
or
$$K(1+\epsilon)\approx K'(1)(1+\epsilon-1)+K(1)$$ and
$$K(1+\epsilon)\approx \frac{\epsilon}{25}+\frac{1}{10}$$