Derivatives of ln y and sin ^-1 (y)

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Make a chain of a function and its inverse:  f^-1(f(x)) = x starts with x and ends with x.
Take the slope using the Chain Rule.   On the right side the slope of x is 1.

Chain Rule:  dx/dy dy/dx = 1   Here this says that df^-1/dy times df/dx equals 1.

So the derivative of f^-1(y)  is  1/ (df/dx)  BUT you have to write df/dx in terms of y.
The derivative of ln y is 1/ (derivative of f = e^x) = 1/e^x.   This is  1/y, a neat slope !
Changing letters is OK :  The derivative of ln x is 1/x.  Watch this video for GRAPHS

Professor Strang's Calculus textbook (1st edition, 1991) is freely available here.

Subtitles are provided through the generous assistance of Jimmy Ren.

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Lecture summary and Practice problems (PDF)

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