# Change in y with change in x

• andyrk
In summary, changing x in y = f(x) from x to x+dx results in a change in y from f(x) to f(x+dx). However, changing y from y to y+dy does not necessarily allow us to determine how x changes. This is because y is defined as a function of x, and not vice-versa. In some cases, if f is invertible, we can approximate the change in x as dx = (1/f'(x)) * dy. However, this method does not always work and may fail if f'(x) = 0.

#### 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?

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

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##.

andyrk said:
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.