Here's the question you clicked on:
Jyqft
If \[y=vx\] becomes \[y=vdx\], then what does \[y=vx^n\] become?
OK, I assume v is a constant?
OK then, the derivative of 'x' is a constant
like you so elegantly show. Now are you familiar with the power rule for derivatives?
Yeah. Derivative of \[x^n\] is \[nx^{n-1}\ But in this case, how can I express that as d(something)?
\[n*x^{n-1}\] Yeah, well, \[dx\] I think can just be looked at as a term for 'change in x'
so what you're actually doing is such: you take the derivative of something relative to something else, so you found that \[y=vx, y'=vdx\] which is \[dy = v * dx \]
so a change in y for a change in x for the original equation can be attributed to v * the change in the x value.
So, y=vx, say v = 2, now when x=2, y = 2*(2) = 4, then when x=3, y = 2*(3) = 6, so the difference in y is 6 - 4 =2, the difference in x is 3-2=1, so when x changed by 1 y changed by 2*the change in x
Now, for the more complex \[y = v * x^{n}\]
now, the change in y with respect to change in x is now time for the derivative
\[dy = v * (n)*(x^{n-1}) * dx\] Use the power rule here
Thanks, that was well-explained.
Nice, is it clear? The cool thing about math is understanding the symbols, but it is easy to let the symbols confuse yourself
so a change in y for a change in x for the original equation can be attributed to v * the change in the x value. That's the key concept for derivatives here, just then you have to find how x changes, and you used the power rule to find how x changes