What is the Taylor series for $g(x) =\frac{ \sinh(-x^{1/2})}{(-x^{1/2})}$, for $x < 0$? 
What is the Taylor series for $$g(x) = \frac{\sinh((-x)^{1/2})}{(-x)^{1/2}}$$, for $x < 0$?

Using the standard Taylor Series:
$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}$$
I substituted in $x = x^{1/2}$, since $x < 0$, it would simply be $x^{1/2}$
getting,
$$\sinh(x^{1/2}) = x^{1/2} + \frac{x^{3/2}}{3!} + \frac{x^{5/2}}{5!} + \frac{x^{7/2}}{7!}$$
Then to get the Taylor series for $\sinh((-x)^{1/2})/((-x)^{1/2})$, would I just divide each term by $x^{1/2}$?
This gives me, $1+\frac{x}{3!}+\frac{x^2}{5!}+\frac{x^3}{7!}$
Is this correct?
Thanks for any help!

As Arturo pointed out in a comment, It has to be $(-x)^{\frac{1}{2}}$ to be defined for $x<0$, then you have:
$$\sinh x = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!}+\dots$$
Substituting $x$ with $(-x)^{\frac{1}{2}}$ we get:
$$\sinh (-x)^{\frac{1}{2}} = (-x)^{\frac{1}{2}} + \frac{({(-x)^{\frac{1}{2}}})^3}{3!} + \frac{({(-x)^{\frac{1}{2}}})^5}{5!} + \frac{({(-x)^{\frac{1}{2}}})^7}{7!}+\dots$$
Dividing by $(-x)^{\frac{1}{2}}$:
$$\frac{\sinh (-x)^{\frac{1}{2}}}{(-x)^{\frac{1}{2}}} = 1 + \frac{({(-x)^{\frac{1}{2}}})^2}{3!} + \frac{({(-x)^{\frac{1}{2}}})^4}{5!} + \frac{({(-x)^{\frac{1}{2}}})^6}{7!}+\dots$$
And after simplification:
$$\frac{\sinh (-x)^{\frac{1}{2}}}{(-x)^{\frac{1}{2}}} = 1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!}+\dots$$
