## Jyqft 3 years ago If \[y=vx\] becomes \[y=vdx\], then what does \[y=vx^n\] become?

1. dpflan

OK, I assume v is a constant?

2. Jyqft

Yup, it is.

3. dpflan

OK then, the derivative of 'x' is a constant

4. dpflan

like you so elegantly show. Now are you familiar with the power rule for derivatives?

5. Jyqft

Yeah. Derivative of \[x^n\] is \[nx^{n-1}\ But in this case, how can I express that as d(something)?

6. dpflan

\[n*x^{n-1}\] Yeah, well, \[dx\] I think can just be looked at as a term for 'change in x'

7. dpflan

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 \[dy = v * dx \]

8. dpflan

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.

9. dpflan

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

10. dpflan

Now, for the more complex \[y = v * x^{n}\]

11. dpflan

now, the change in y with respect to change in x is now time for the derivative

12. dpflan

\[dy = v * (n)*(x^{n-1}) * dx\] Use the power rule here

13. Jyqft

Thanks, that was well-explained.

14. dpflan

Nice, is it clear? The cool thing about math is understanding the symbols, but it is easy to let the symbols confuse yourself

15. dpflan

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