Posted by Anonymous on Thursday, May 12, 2011 at 12:54am.
Let f(x) = y = x^3 + x
Since a cubic function is involved, I cannot write x(y) or g(x) explicitly.
y = 2 when x = 1. That is why g(2) = 1
dy/dx = 3x^2 + 1
When x = 1, which is when the inverse function g =2, then dy/dx = 4
The derivative of g(x) at that point is dx/dy = 1/(dy/dx) = 1/4
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