Validity of sinh(x^2): Explained

[JamesGoh]

Is the following statement valid?

$sinh{x^2}=\frac{e^{x^2}-e^{x^2}}{2}$

Reason I ask cause I know that $sinh{x}=\frac{e^{x}-e^{x}}{2}$, so i assume the same rules can apply

 

[Char. Limit]

If by chance you meant $sinh(x^{2}) = \frac{e^{x^2} - e^{-x^2}}{2}$, then yes, that is the case.

Simply asking because you forgot to include that negative sign in the second exponent.

 

[HakimPhilo]

As stated by Char. Limit the correct statement is $\sinh \color{blue}z\equiv\dfrac{e^{\color{blue}z}-e^{-\color{blue}z}}2$ then substituting $z\rightsquigarrow x^2$ we get: $$\sinh x^2=\frac{e^{x^2}-e^{-x^2}}{2}.$$