Question

dy/dx (x-1)^1/2

Answer 1

so the answer is 1/2(x-1)^-1/2 ?

Answer 2

vat do u need to find actually?

Answer 3

i need to find the differentiation of (x-1)^1/2

Answer 4

then that's the answer.

Answer 5

or you looknig for y?
\[y=\int\limits (x-1)^{1/2}dx =2/3(x-1)^{3/2} +C\]

Answer 6

bt if u hav (2x-1)^1/2
dy/dx=1/2(2x-1)^-1/2*2=(2x-1)^-1/2

Answer 7

\[\int\limits_{}^{} x \sqrt{x-1} dx\] the initial question was this. but i'm confused, i don't know how to differentiate (x-1)^1/2

Answer 8

dx is not inside the sq root

Answer 9

\[dx \over 2\sqrt{x-1}\]

Answer 10

thats diferential of (x-1)^1/2

Answer 11

the answer i have here is 2/5(x-1)^5/2 + 2/3(x-1)^3/2 + C

Answer 12

this is for the integral

Answer 13

i substituted x-1 for u. and i got stuck at \[\int\limits_{}^{}x u ^{1/2}\]
after that, i do not know how to carry on. i'm confused on the substitution part for this. is it possible for someone to show me the steps? so i can understand it. tutorials i have read says that i have to make the x disappear, but i do not know how.

Answer 14

x-1=u
x=u+1

Answer 15

okay. the whole question is this:
Determine the integral \[\int\limits_{}^{}x \sqrt{x-1} dx\] bu using the substitution u = x-1

Answer 16

\[\int\limits (u+1)\sqrt{u}du = \int\limits \sqrt{u ^{3}}du + \int\limits \sqrt{u}du\]

Answer 17

i'm sorry myko, i have no idea what you're typing. i have lost touch with maths for 2 years.

Answer 18

when you make substitution:
x-1 = u
and
x=u+1
you get the integral i typed befor

Answer 19

so i am making the equation to be in terms of u? i don't understand the 2nd part of the answer. i'm so sorry.

Answer 20

you substitute the x-1 by u
and x by u+1

Answer 21

okay, i understand what you're saying. but how is it \[u \times \sqrt{u} = \sqrt{u ^{3}}\]

Answer 22

oh don't have to reply to my previous question. i got it

Answer 23

thanks alot myko, i got the answer.