anonymous
  • anonymous
If \[y=vx\] becomes \[y=vdx\], then what does \[y=vx^n\] become?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
OK, I assume v is a constant?
anonymous
  • anonymous
Yup, it is.
anonymous
  • anonymous
OK then, the derivative of 'x' is a constant

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
like you so elegantly show. Now are you familiar with the power rule for derivatives?
anonymous
  • anonymous
Yeah. Derivative of \[x^n\] is \[nx^{n-1}\ But in this case, how can I express that as d(something)?
anonymous
  • anonymous
\[n*x^{n-1}\] Yeah, well, \[dx\] I think can just be looked at as a term for 'change in x'
anonymous
  • anonymous
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 \[dy = v * dx \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
Now, for the more complex \[y = v * x^{n}\]
anonymous
  • anonymous
now, the change in y with respect to change in x is now time for the derivative
anonymous
  • anonymous
\[dy = v * (n)*(x^{n-1}) * dx\] Use the power rule here
anonymous
  • anonymous
Thanks, that was well-explained.
anonymous
  • anonymous
Nice, is it clear? The cool thing about math is understanding the symbols, but it is easy to let the symbols confuse yourself
anonymous
  • anonymous
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.