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1 solution
For any differentiable function f ( x ) with an inverse function g ( x ) where f ′ ( g ( x 0 ) ) exists and is not zero, g ′ ( x 0 ) = f ′ ( g ( x 0 ) ) 1 Assuming f ′ ( g ( x ) ) = 0 and f ( g ( x ) ) is differentiable at given x , g ′ ( x ) = f ′ ( g ( x ) ) 1 = 1 + g ( x ) 5 1 1 = 1 + g ( x ) 5