## mathcalculus 2 years ago help this simply derivative question: f(x)= sqrt(5x)

1. mathcalculus

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

how come 5 is also a square root?

3. mathcalculus

what step am i missing?

4. wio

$\sqrt{5x} = \sqrt{5}\sqrt{x}$

5. wio

So the derivative is $\sqrt{5}\frac{1}{2\sqrt{x}}$

6. wio

@mathcalculus Make sense?

7. KeithAfasCalcLover

\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

OOPS lol got cut off. wait...

9. KeithAfasCalcLover

\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

@wio so you can't re-write 5 as 5^(1/2)?

11. KeithAfasCalcLover

Well, you can! But it would still be a constant.

12. mathcalculus

?