# How do you find (f^-1)'(a) if f(x)= x/sqrt{x-3} and a = 4?

Oct 6, 2016

The function is not bijective, So, the bi-valued inverse is not differentiable. See the explanation..

#### Explanation:

y = f(x) is real for x >=3.

$y ' = \frac{1}{x - 3} - \frac{x}{2 {\left(x - 3\right)}^{\frac{3}{2}}} = \frac{x - 6}{2 {\left(x - 3\right)}^{\frac{3}{2}}}$.

Inversely, bi-valued $x = \frac{y \left(y \pm \sqrt{{y}^{2} - 12}\right)}{2}$, with $| y | \ge 2 \sqrt{3}$

However, assigning y = 4, x = 4 or 12.for the two inverse forms.