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Suppose f is a one-to-one, differentiable function and its inverse function f^−1 is also differentiable. One can show, using implicit differentiation (do it!), that

(f^−1)′(x)=1 / f′(f^−1(x))

Find (f^−1)′(−6) if f(−1)=−6 and f′(−1)=3/5.

(f^−1)′(−6)=

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  1. you have g(x) = f^-1(x)
    Since f(-1) = -6, g(-6) = -1
    (f^-1(-6))' = 1/g'(-1) = 1/(3/5) = 5/3

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    oobleck

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