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mathcalculus

  • 2 years ago

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

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  1. mathcalculus
    • 2 years ago
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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)

  2. mathcalculus
    • 2 years ago
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    how come 5 is also a square root?

  3. mathcalculus
    • 2 years ago
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    what step am i missing?

  4. wio
    • 2 years ago
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    \[ \sqrt{5x} = \sqrt{5}\sqrt{x} \]

  5. wio
    • 2 years ago
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    So the derivative is \[ \sqrt{5}\frac{1}{2\sqrt{x}} \]

  6. wio
    • 2 years ago
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    @mathcalculus Make sense?

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

  8. KeithAfasCalcLover
    • 2 years ago
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    OOPS lol got cut off. wait...

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

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

  11. KeithAfasCalcLover
    • 2 years ago
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    Well, you can! But it would still be a constant.

  12. mathcalculus
    • 2 years ago
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    ?

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