So, I need to find (F^-1)'(a) of

f(x) = 3 x^2 + tan(πx/2) , -1< x < 1, a=3

I have worked the problem to this point

(f^-1)'(3) = 1/2x + sec^2(πx/2)π/2

= 1/2(0) + sec^2(π(0)/2)π/2

but my book and my professor tell me that the answer is 2/π, but I don't see how that can be. Shouldn't the answer be 1/0 because the 0's in the denominator give you 0 + 0, which is 0?