Change in y with change in x

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Suppose we change x in y = f(x) from x to x+dx, then y changes from f(x) to f(x+dx). But suppose if we change y from y to y+dy, then can we determine how x changes? Why or why not? Is it because y is a function of x and not vice-versa?
 

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  • #2
lurflurf
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sure if f is invertible we will have x change to $$\mathrm{f}^{-1}(\mathrm{f}(x)+\mathrm{dy})$$
 
  • #3
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Take the sinus function: You cannot always determine what dx was from dy. However, if the ds are small quantities and f is differentiable, in many cases you can take the approximation ##y(x+dx)\approx l(x+dx)=f(x)+f'(x)\cdot dx##. You also know you started with ##y=f(x)##, so in linear approximation ##dy=f'(x) \cdot dx## and ##dx=\frac{1}{f'(x)}\cdot dy##. Of course, this does not always work, for example it does not work if ##f'(x)=0##.
 
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Svein
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Suppose we change x in y = f(x) from x to x+dx, then y changes from f(x) to f(x+dx). But suppose if we change y from y to y+dy, then can we determine how x changes? Why or why not? Is it because y is a function of x and not vice-versa?
The way you defined y (as f(x)) means that y depends on x and you are supposed to take the result of f(x) and put it into y. Changing the y without changing the x is sort of violating the rules.

But - given y, there may be an x such that y = f(x). And for those y it may be possible to talk about an inverse function g, defined by g(f(x)) = x. See a suitable book on mathematical analysis.
 

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