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mathcalculusBest ResponseYou've already chosen the best response.0
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)
 6 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
how come 5 is also a square root?
 6 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
what step am i missing?
 6 months ago

wioBest ResponseYou've already chosen the best response.1
\[ \sqrt{5x} = \sqrt{5}\sqrt{x} \]
 6 months ago

wioBest ResponseYou've already chosen the best response.1
So the derivative is \[ \sqrt{5}\frac{1}{2\sqrt{x}} \]
 6 months ago

KeithAfasCalcLoverBest ResponseYou've already chosen the best response.0
\[\eqalign{ &f(x)=\sqrt{5x} \\ &\\ &\\ &\frac{d}{dx}f(x)=\frac{d}{dx}\sqrt{5x}=\frac{d}{dx}\sqrt{5}\sqrt{x}=\sqrt{5}\frac{d}{dx}\sqrt{x}=\sqrt{5}\frac{1}{2\sqrt{x}}=\frac{\sqrt{5}}{2\sqrt{x}}=\frac{\sqrt{5}}{\sqrt{4x}}=\sqrt{\frac{5}{4x}} \\ }\]
 6 months ago

KeithAfasCalcLoverBest ResponseYou've already chosen the best response.0
OOPS lol got cut off. wait...
 6 months ago

KeithAfasCalcLoverBest ResponseYou've already chosen the best response.0
\[\eqalign{ &f(x)=\sqrt{5x} \\ &\\ &\\ &\eqalign{\frac{d}{dx}f(x) &=\frac{d}{dx}\sqrt{5x} \\ &=\frac{d}{dx}\sqrt{5}\sqrt{x} \\ &=\sqrt{5}\frac{d}{dx}\sqrt{x} \\ &=\sqrt{5}\frac{1}{2\sqrt{x}} \\ &=\frac{\sqrt{5}}{2\sqrt{x}} \\ &=\frac{\sqrt{5}}{\sqrt{4x}} \\ &=\sqrt{\frac{5}{4x}} \\} \\ }\]
 6 months ago

mathcalculusBest ResponseYou've already chosen the best response.0
@wio so you can't rewrite 5 as 5^(1/2)?
 6 months ago

KeithAfasCalcLoverBest ResponseYou've already chosen the best response.0
Well, you can! But it would still be a constant.
 6 months ago
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