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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OK, I assume v is a constant?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OK then, the derivative of 'x' is a constant

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0like you so elegantly show. Now are you familiar with the power rule for derivatives?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah. Derivative of \[x^n\] is \[nx^{n1}\ But in this case, how can I express that as d(something)?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[n*x^{n1}\] Yeah, well, \[dx\] I think can just be looked at as a term for 'change in x'

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0So, 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 32=1, so when x changed by 1 y changed by 2*the change in x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, for the more complex \[y = v * x^{n}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now, the change in y with respect to change in x is now time for the derivative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[dy = v * (n)*(x^{n1}) * dx\] Use the power rule here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks, that was wellexplained.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Nice, is it clear? The cool thing about math is understanding the symbols, but it is easy to let the symbols confuse yourself

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
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