# Derivative function of inverse function

Calculus Level pending

Let $$g(x)$$ be the derivative of the inverse function of $$f(x) = \tan^2x$$ for $$0 \lt x \lt \frac{\pi}{2} .$$ If the minimum value of $$h(x)=\frac{-g'(x)}{\left(g(x)\right)^2}$$ is $$a,$$ which is attained at $$x=b,$$ then what is $$a^{2} + 60b?$$

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