Change in y with change in x

[andyrk]

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?

 

[lurflurf]

sure if f is invertible we will have x change to $$\mathrm{f}^{-1}(\mathrm{f}(x)+\mathrm{dy})$$

 

[DarthMatter]

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 $f'(x)=0$.

 

[Svein]

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.