Derivatives of Inverse Functions

msinc210

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Oct 30, 2013
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Hi guys--I've been having trouble with this topic. My teacher wants us to use the relationship dy/dx = 1/(dx/dy) to help solve derivatives of inverse functions.


For example for function f(x) = 3x^2 + 4y^2 = 12, find d/dx of f^-1(x) at x = 1

Basically how we've been solving this is by getting dx/dy of the regular (non inverse function) and then plugging in y = 1. I don't understand at all the reasoning behind this. What is the significance of the getting reciprocal of a dy/dx (aka dx/dy) as opposed to just dy/dx of the preliminary function, and then plugging in y = 1 instead of x = 1 to find the formula for the inverse function?


Also for Inverse Trig Functions, differentiating them seems completely unrelated. For example, to get derivative of y = sin^-1(x), we have to switch x and y and get the inverse of the inverse trig function (I think?) and end up with x = sin(y). We then take dy/dx of that new function (which confuses me because, as above, we used dx/dy) to get dy/dx = 1/cosy. I then understand the final steps to get dy/dx = 1/rad(1-x^2).

I'm sorry if this post is confusing. I'll be on here if you have any questions.
 
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