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

Question

anonymous · Answer

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)

anonymous · Answer

how come 5 is also a square root?

anonymous · Answer

what step am i missing?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

@mathcalculus Make sense?

anonymous · Answer

$$\eqalign{
&f(x)=\sqrt{5x} \
&\
&\
&\frac{d}{dx}f(x)=\frac{d}{dx}\sqrt{5x}=\frac{d}{dx}\sqrt{5}\sqrt{x}=\sqrt{5}\frac{d}{dx}\sqrt{x}=\sqrt{5}\frac{1}{2\sqrt{x}}=\frac{\sqrt{5}}{2\sqrt{x}}=\frac{\sqrt{5}}{\sqrt{4x}}=\sqrt{\frac{5}{4x}} \
}$$

anonymous · Answer

OOPS lol got cut off. wait...

anonymous · Answer

$$\eqalign{
&f(x)=\sqrt{5x} \
&\
&\
&\eqalign{\frac{d}{dx}f(x)
&=\frac{d}{dx}\sqrt{5x} \
&=\frac{d}{dx}\sqrt{5}\sqrt{x} \
&=\sqrt{5}\frac{d}{dx}\sqrt{x} \
&=\sqrt{5}\frac{1}{2\sqrt{x}} \
&=\frac{\sqrt{5}}{2\sqrt{x}} \
&=\frac{\sqrt{5}}{\sqrt{4x}} \
&=\sqrt{\frac{5}{4x}} \} \
}$$

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

?