So I understand how to find the derivative of $f(x)=x^{1/2}$ using the power rule.
I also know how to find it using the limit. $f'(x) = \lim_{h \to 0} \frac{(x+h)^{1/2} - x^{1/2}}{h}$
You could replace h and f' in that formula with dx and df/dx to find a formula like this:
$\frac{df}{dx} = \lim_{dx \to 0} \frac{(x+dx)^{1/2} - x^{1/2}}{dx}$
That kind of looks like what I want, but eventually I'm trying to get to this formula I've seen around
$df = (x+dx)^{1/2} - x^{1/2}$
Then I want to use that to get to df = something*dx
I can't figure out how to isolate the dx, which sucks.
An example of the logic I want is in the following proof about $f=x^2$
$df = (x+dx)^{2}-x^2$
$df = x^2+2x*dx+(dx)^2-x^2$
$df = 2x*dx+(dx)^2$
Then you can ignore an infinitesimal df times itself as inconsequential.
So $df = 2x*dx$
Does anyone even know what I mean when I'm asking this question? I am confused myself, but I want to understand the df = something * dx way of doing things
Think it will be useful for linear algebra when x is a matrix
Hint: $df=(x+dx)^{1/2}−x^{1/2} \implies df=x^{1/2}(1+\frac{dx}{x})^{1/2}−x^{1/2} \implies df=x^{1/2}(1+\frac{dx}{2x})−x^{1/2}$$\implies df=\frac{dx}{2\sqrt{x}} $ (Using the approximation $(1+x)^n=1+nx$ when $x$ is very small compared to $1$.)