Question

Show steps on how to find derivative of
y = (x^2 - 2x)^(1/2)

Answer 1

\[y = \sqrt{x^2 -2x}\]
\[y' =\]

Answer 2

Use the product rule, I think that will work.

Answer 3

Take the derivative of the outside function, then multiply it by the inside

Answer 4

So:
\[\LARGE (x^2-2x)^{1/2}=\frac{1}{2(x^2-2x)}*2x-2\]
Make some sense?

Answer 5

Sorry that should be:
\[\sqrt{x^2-2x}\]

Answer 6

I understand where you get 2x-2 from, but why is there a fraction?

Answer 7

Sham, catch up bro:
\[\LARGE x^{1/2}=\frac{1}{2}*x^{-1/2}\]

Answer 8

@shamil98 understand?

Answer 9

@NicholasWong752 Do YOU understand?

Answer 10

Power rule followed by chain rule.
Or,
Let t = x^2 - 2x
y = t^1/2
dy/dx = (dy/dt)(dt/dx)
dy/dt = (1/2t^(-1/2) = 1/(2(t^1/2)) = 1/(2(x^2 - 2x)^1/2)
dt/dx = 2x - 2
dy/dx = (2x - 2)/(2(x^2 - 2x)^1/2) = (x - 1)/(x^2 - 2x)^1/2

Answer 11

Where are you guys getting that (x-1)?

Answer 12

\[\sqrt{x^2-2x} = \frac{ 1 }{ 2(x^2 -2x) } \times 2x -2\]
\[\sqrt{x^2-2x} = \frac{ (x-1) }{ (x^2-2x) } \]

Answer 13

2(x-1)/2(x^2-2x)
the twos cancel out :P

Answer 14

I understand why there's fraction, but I got stuck at that point

Answer 15

Oh, right I didn't simplify..

Answer 16

x-1/√(x^2-2x)
right?

Answer 17

Need a total summary of everything we did? @NicholasWong752

Answer 18

Yes sham, nice job kiddo