## sdavenko 3 years ago what is the derivative of x(2x-3)^1/2?

1. dpaInc

(2x-3)^(1/2) + x(2x-3)^(-1/2)

2. sdavenko

what about the 2 that is taken from the inside of the function? and thats what i have already but i dont know how to simplify it further

3. arcticf0x

That 2 has been cancelled by 1/2 from the exponent. Do you have some answer you have to simplify for?

4. sdavenko

oh thanks butthis isnt an integration problem

5. sdavenko

the other options are a) 3x-3/ sqrt2x-3 b.) x/sqrt 2x-3 c.) 1/sqrt 2x-3 d.)-x+3/ sqrt 2x-3 e.)5x-6/2 sqrt 2x-3

6. arcticf0x

Simply add the terms, you will get 3x-3/(2x-3)^(1/2)...Know how?

7. sdavenko

no the final step im on is (2x-3)^1/2 +(2x)(1/2)(2x-3)^-1/2

8. arcticf0x

Ok, look, we have, (2x-3)^(1/2) +x(2x-3)^(-1/2) => [(2x-3)^(1/2) * (2x-3)^(1/2) + x ]/(2x-3)^(-1/2) =>(2x-3+x)/(2x-3)^(-1/2) =>3x-3/(2x-3)^(1/2)

9. sdavenko

ah i feel th feel like i have made this problem harder than it neeeds to be but, why did you multiply (2x-3)^(1/2) by (2x-3)^(1/2) in the second step?

10. arcticf0x

Its the way you add surds, multiply the denominator and join them.

11. sdavenko

ok thanks! I really appreciate you're patience

12. arcticf0x

I wish i could have done that in proper latex so it'd have been clear before.

13. sdavenko

where are you from? what country?