Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnecessarily complicated? If there are any other methods to finding the minimum value I am open to those too.

A more efficient answer than my previous approach:
Starting in the same way:
Since $\sin\left(\frac{\pi}{2}-x\right)=\cos(x)$, $f(x)$ is symmetric around $x=\frac{\pi}{4}$.  This implies that $x=\frac{\pi}{4}$ is a critical point.
Writing $f(x)=(1+\sin(x)^{-6})(1+\cos(x)^{-6})$, we find that
$$
f'(x)=-6\sin(x)^{-7}\cos(x)(1+\cos(x)^{-6})+6\cos(x)^{-7}\sin(x)(1+\sin(x)^{-6}).
$$
This derivative can be factored as
$$
f'(x)=6\sin(x)^{-7}\cos(x)^{-7}(-\cos(x)^8-\cos(x)^2+\sin(x)^8+\sin(x)^2).
$$
Since $\cos(x)>\sin(x)$ for $0<x<\frac{\pi}{4}$ and $\cos(x)<\sin(x)$ for $\frac{\pi}{4}<x<\frac{\pi}{2}$, the cosine terms dominate for $0<x<\frac{\pi}{4}$ and the sine terms dominate for $\frac{\pi}{4}<x<\frac{\pi}{2}$.
Therefore, the derivative is negative for $0<x<\frac{\pi}{4}$ and positive for $\frac{\pi}{4}<x<\frac{\pi}{2}$.  Hence, $x=\frac{\pi}{4}$ is the global minimum.
