If fff has inverse f−1f^{-1}f−1 and f(2)=5f(2) = 5f(2)=5, f′(2)=4f'(2) = 4f′(2)=4, then (f−1)′(5)=(f^{-1})'(5) =(f−1)′(5)=A14\tfrac{1}{4}41check_circleB444C15\tfrac{1}{5}51D555Explanation(f−1)′(b)=1/f′(a)(f^{-1})'(b) = 1/f'(a)(f−1)′(b)=1/f′(a) where f(a)=bf(a) = bf(a)=b. Here 1/f′(2)=1/41/f'(2) = 1/41/f′(2)=1/4.