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July 23, 2014

July 23, 2014

Posted by **Anonymous** on Thursday, May 12, 2011 at 12:54am.

- calculus -
**drwls**, Thursday, May 12, 2011 at 9:33amLet 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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