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Jyqft

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

  • one year ago
  • one year ago

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  1. dpflan
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    OK, I assume v is a constant?

    • one year ago
  2. Jyqft
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    Yup, it is.

    • one year ago
  3. dpflan
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    OK then, the derivative of 'x' is a constant

    • one year ago
  4. dpflan
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    like you so elegantly show. Now are you familiar with the power rule for derivatives?

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

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

    • one year ago
  7. dpflan
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    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
  8. dpflan
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    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
  9. dpflan
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    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

    • one year ago
  10. dpflan
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    Now, for the more complex \[y = v * x^{n}\]

    • one year ago
  11. dpflan
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    now, the change in y with respect to change in x is now time for the derivative

    • one year ago
  12. dpflan
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    \[dy = v * (n)*(x^{n-1}) * dx\] Use the power rule here

    • one year ago
  13. Jyqft
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    Thanks, that was well-explained.

    • one year ago
  14. dpflan
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    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
  15. dpflan
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    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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