y= ln(x+3) /2

2y = ln(x+3)

x+3= e^(2y)

x= e^(2y)-3

f^-1(x) = e^(2x)-3

f^-1(0) = e^(2*0) -3 =1-3 = -2

d/dx [f^-1(0)] = 1/ f'(f^-1(0)) = 1/f'(-2)

f'(x) = (1/2)(1/(x+3))

f'(-2) = (1/2) (1/(-2+3)) = 1/2

d/dx [f^-1(0)] = 1/ f'(f^-1(0)) = 1/(1/2)

= 2

An alternate approach without actually working out f^(-1)(x) except at x = 0:

f^(-1)(x) at "x" = 0 means finding the original x value when ln(x + 3)/2 = 0 or ln(x + 3) = 0 or x + 3 = 1, so x = -2.

d/dx[ f^(-1)(x) ] = 1/f'(x), That is, dx/dy = 1/(dy/dx). We'll work out f'(x) at that point and take its reciprocal.

f(x) = (1/2) ln(x + 3)

f'(x) = (1/2) * 1/(-2 + 3). At x = 0, this is (1/2) * (1/1) = 1/2. And so 1/f'(x) = 2.

The other answers gave the general expression 2 e^(2x), which you can see when x = 0 is equal to 2 * e^0 = 2.

x = ln(y + 3)/2;

2x = ln(y + 3);

e^(2x) = y + 3

y = ((e^(2x)) - 3;

y' = 2(e^(2x));

y'(0) = 2*(e^(2*0)) = 2;

Let g(x) = f^(-1)(x), then

f(g(x)) = ln(g(x) + 3)/2 = x

ln(g(x) + 3) = 2x

g(x) + 3 = e^(2x)

g(x) = e^(2x) - 3

g'(x) = 2e^(2x)