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mathcalculus Group Title

help this simply derivative question: f(x)= sqrt(5x)

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    i rewrote it: (5x)^1/2 then multiplied and take away one from the exponent: (5x/2) ^-1/2 then i move the negative exponent in the denominator: 5x/2 sqrt x)

    • one year ago
  2. mathcalculus Group Title
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    how come 5 is also a square root?

    • one year ago
  3. mathcalculus Group Title
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    what step am i missing?

    • one year ago
  4. wio Group Title
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    \[ \sqrt{5x} = \sqrt{5}\sqrt{x} \]

    • one year ago
  5. wio Group Title
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    So the derivative is \[ \sqrt{5}\frac{1}{2\sqrt{x}} \]

    • one year ago
  6. wio Group Title
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    @mathcalculus Make sense?

    • one year ago
  7. KeithAfasCalcLover Group Title
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    \[\eqalign{ &f(x)=\sqrt{5x} \\ &\\ &\\ &\frac{d}{dx}f(x)=\frac{d}{dx}\sqrt{5x}=\frac{d}{dx}\sqrt{5}\sqrt{x}=\sqrt{5}\frac{d}{dx}\sqrt{x}=\sqrt{5}\frac{1}{2\sqrt{x}}=\frac{\sqrt{5}}{2\sqrt{x}}=\frac{\sqrt{5}}{\sqrt{4x}}=\sqrt{\frac{5}{4x}} \\ }\]

    • one year ago
  8. KeithAfasCalcLover Group Title
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    OOPS lol got cut off. wait...

    • one year ago
  9. KeithAfasCalcLover Group Title
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    \[\eqalign{ &f(x)=\sqrt{5x} \\ &\\ &\\ &\eqalign{\frac{d}{dx}f(x) &=\frac{d}{dx}\sqrt{5x} \\ &=\frac{d}{dx}\sqrt{5}\sqrt{x} \\ &=\sqrt{5}\frac{d}{dx}\sqrt{x} \\ &=\sqrt{5}\frac{1}{2\sqrt{x}} \\ &=\frac{\sqrt{5}}{2\sqrt{x}} \\ &=\frac{\sqrt{5}}{\sqrt{4x}} \\ &=\sqrt{\frac{5}{4x}} \\} \\ }\]

    • one year ago
  10. mathcalculus Group Title
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    @wio so you can't re-write 5 as 5^(1/2)?

    • one year ago
  11. KeithAfasCalcLover Group Title
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    Well, you can! But it would still be a constant.

    • one year ago
  12. mathcalculus Group Title
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    ?

    • one year ago
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