Question

find the derivative of:
sqrt(2x)

Book says f'(x)= sqrt(2)/(2*sqrt(x))

I can't get that.

Answer 1

its 2x^.5 so \[(.5)(2)x^(.5-1) = x^(-.5)\]

Answer 2

then, x^-.5 = \[1/sqrtx\]

Answer 3

idk y your book says that

Answer 4

That's what I've been getting, too. I even tried deriving via the differential and couldn't come to the book's answer.

Answer 5

*differential equation.

Answer 6

is there another step in the instructions? but thats the derivative fersure

Answer 7

In Exercise 1 through 28, differentiate the given function. Simplify your answers.

\[y=\sqrt{2x}\]

Answer 8

oh. then idk what to tell you. thats the answer tho. just one problem dude. just move on. YOU know your right.

Answer 9

one way to see the answer:
\[\sqrt{2x}= \sqrt{2}x^{\frac{1}{2}}\]
and the derivative is
\[\sqrt{2} \cdot \frac{1}{2}x^{-\frac{1}{2}}\]

Answer 10

In other words
\[\sqrt{2}\frac{d}{dx}x^{\frac{1}{2}}= \sqrt{2}\frac{1}{2}x^{-\frac{1}{2}}\]

Answer 11

^^ nice

Answer 12

The other way is to use the chain rule

Answer 13

\[\frac{d}{dx}(2x)^{\frac{1}{2}}= \frac{1}{2}(2x)^{-\frac{1}{2}}\cdot 2=\frac{1}{\sqrt{2}{\sqrt{x}}}= \frac{\sqrt{2}}{2\sqrt{x}}\]

Answer 14

I wasn't looking at \[\sqrt{2}\] as a constant. Thank you!