Clarification of the proof of Corollary V.4.5 in Bhatia "Matrix Analysis"

The proof of the corollary is just noting that the given non-constant operator monotone function on $(-1,1)$ satisfies $f^{\prime }(0)\ne 0$. But the book just states that this is true because $f$ is monotone. This reason alone is insufficient.

To fix the proof, we apply Theorem V.3.4 to the matrix $A=diag(t,0)$ with $t\ne 0$. Then the following matrix is positive semi-definite.
$$f^{[1]}(A)=\left(\begin{array}{cc}{f}^{\prime }(t)& \frac{f(t)-f(0)}{t}\\ \frac{f(t)-f(0)}{t}& f^{\prime }(0)\end{array}\right)$$
As $f$ is non-constant, there is some $t\ne 0$ such that $f(t)\ne 0$. By positive semi-definiteness, $f^{\prime }(t)\ge 0$ and $\det (f^{[1]}(A))\ge 0$. Hence
$$f^{\prime }(t)f^{\prime }(0)-{\left(\frac{f(t)-f(0)}{t}\right)}^2\ge 0$$
$$f^{\prime }(t)f^{\prime }(0)\ge {\left(\frac{f(t)-f(0)}{t}\right)}^2>0$$
This implies that $f^{\prime }(0)>0$.