Question

find the first derivative sqrt[x^2-sqrt(2x-3)]

Answer 1

chain rule

Answer 2

try it

Answer 3

is this right? sqrt[x^2-(1/2(2x-3)^-1/2 * 2)]?

Answer 4

We can let \(f(x) = \sqrt{x}\) and \(g(x) = x^2-\sqrt{2x-3}\), so we have \(f(g(x))\) to apply chain rule

Answer 5

\(\dfrac{d}{dx}f(~g(x)~) = f'(~g(x)~)~g'(x)\)

Answer 6

or this 1/2*sqrt[x^2-sqrt(2x-3)]* d/dx (x2-sqrt 2x-3)

Answer 7

yes I think... hard to read O_o

If you mean \(\Large \dfrac{1}{2\sqrt{x^2-\sqrt{2x-3}}}\cdot\dfrac{d}{dx}\left(x^2-\sqrt{2x-3}\right)\), then you are right

Answer 8

so finish derivative and put it in numerator, and you are done. i think you can simplify, but i'm sure teacher don't want you to so that

Answer 9

is that clear? @wakakaka

Answer 10

yeah thanks a lot

Answer 11

welcome