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mathcalculus
help this simply derivative question: f(x)= sqrt(5x)
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)
how come 5 is also a square root?
what step am i missing?
\[ \sqrt{5x} = \sqrt{5}\sqrt{x} \]
So the derivative is \[ \sqrt{5}\frac{1}{2\sqrt{x}} \]
\[\eqalign{ &f(x)=\sqrt{5x} \\ &\\ &\\ &\frac{d}{dx}f(x)=\frac{d}{dx}\sqrt{5x}=\frac{d}{dx}\sqrt{5}\sqrt{x}=\sqrt{5}\frac{d}{dx}\sqrt{x}=\sqrt{5}\frac{1}{2\sqrt{x}}=\frac{\sqrt{5}}{2\sqrt{x}}=\frac{\sqrt{5}}{\sqrt{4x}}=\sqrt{\frac{5}{4x}} \\ }\]
OOPS lol got cut off. wait...
\[\eqalign{ &f(x)=\sqrt{5x} \\ &\\ &\\ &\eqalign{\frac{d}{dx}f(x) &=\frac{d}{dx}\sqrt{5x} \\ &=\frac{d}{dx}\sqrt{5}\sqrt{x} \\ &=\sqrt{5}\frac{d}{dx}\sqrt{x} \\ &=\sqrt{5}\frac{1}{2\sqrt{x}} \\ &=\frac{\sqrt{5}}{2\sqrt{x}} \\ &=\frac{\sqrt{5}}{\sqrt{4x}} \\ &=\sqrt{\frac{5}{4x}} \\} \\ }\]
@wio so you can't re-write 5 as 5^(1/2)?
Well, you can! But it would still be a constant.