if f is one - one onto and differentiablw on R, then, (f−1(6))′=1/f′(6)(f^{-1}(6))^{'}=1/f^{'}(6)(f−1(6))′=1/f′(6) is true because (f−1(yo))′=1/f′(xo)(f^{-1}(y_o))^{'}=1/f^{'}(x_o)(f−1(yo))′=1/f′(xo) for f(x)=xf(x)=xf(x)=x .
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