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dpflanBest ResponseYou've already chosen the best response.2
OK, I assume v is a constant?
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
OK then, the derivative of 'x' is a constant
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
like you so elegantly show. Now are you familiar with the power rule for derivatives?
 one year ago

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

dpflanBest ResponseYou've already chosen the best response.2
\[n*x^{n1}\] Yeah, well, \[dx\] I think can just be looked at as a term for 'change in x'
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
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 \[dy = v * dx \]
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
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.
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
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 32=1, so when x changed by 1 y changed by 2*the change in x
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
Now, for the more complex \[y = v * x^{n}\]
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
now, the change in y with respect to change in x is now time for the derivative
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
\[dy = v * (n)*(x^{n1}) * dx\] Use the power rule here
 one year ago

JyqftBest ResponseYou've already chosen the best response.0
Thanks, that was wellexplained.
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
Nice, is it clear? The cool thing about math is understanding the symbols, but it is easy to let the symbols confuse yourself
 one year ago

dpflanBest ResponseYou've already chosen the best response.2
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
 one year ago
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