Question

I cant seem to figure this problem out :(
find y' when y=2√x-1/2√x

Answer 1

you mean that? y = (2*sqrt(x-1))/(2*sqrt(x))

Answer 2

no i mean 2sqrtx - (1/2sqrtx)

Answer 3

Ok, y = 2*sqrt(x) - 1/(2*sqrt(x))
y' = 2/(2*sqrt(x)) - (-2/(2*sqrt(x)))/(2*sqrt(x))^2
y' = 1/sqrt(x) + (1/sqrt(x))/(4*x)
y' = 1/sqrt(x) + 1/(4*x*sqrt(x))

Answer 4

just remember that f(x) = sqrt(x) then f'(x) = 1/(2*sqrt(x)) .
And h(x) = f(x)/g(x) then h'(x) = (f'(x)*g(x)-f(x)*g'(x))/(g(x))^2

Answer 5

omg thank you sooo much. I've been struggling for hours on this problem. :)

Answer 6

No problem ;)

Answer 7

wait im a little confused. How did you get 2 on the numerator for 2/sqrt*x ?

Answer 8

OK, let's differentiate f(x) = 1/(2*sqrt(x))

Answer 9

f'(x) = (0 - 1*(2/(2*sqrt(x))))/(2*sqrt(x))^2

Answer 10

see if you understand now

Answer 11

Hmm it's kind of hard for me to understand because we can't actually write it out like the way it would look on paper u know? but i'll just look at the work u gave me and hopefully i'll understand eventually. Thank you so much though:)