Question

find dy/dx for (sqrt(x)-1)/(sqrt(x) +1)

Answer 1

if possible could you assist me @ganeshie8 i would appreciate it

Answer 2

http://www.wolframalpha.com/input/?i=derivative+of+y%3D%28x+%5E%7B1%2F2%7D-1%29%2F%28x%5E%7B1%2F2%7D%2B1%29

Answer 3

could you explain how you got that answer for the derivative of [(sqrt{x}-1)/(sqrt{x}+1)

Answer 4

look the power rule says: func) x^n = Deri) nx^n−1

Answer 5

nx^n-1

Answer 6

i understand that but could you explain it in steps as i am confused where to apply the power rule for this equation

Answer 7

So \[\sqrt x\] can be written as \[\huge x^{1\over2} \]

Answer 8

yes i understand that

Answer 9

So you differentiate it, and show up!

Answer 10

thank you for your time and effort @Rohangrr i understand now

Answer 11

Thanks, hope it helped. Any more help needed when I'm offline just mail @ rcdrohan@gmail.com

Answer 12

Arigato ozaimas ~

Answer 13

For a function of the form \[y(x)=\frac{f(x)}{g(x)}\]the quotient rule says\[y'(x)=\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}\]

\[y(x) = \frac{\sqrt x-1}{\sqrt x +1}\]\[\frac{\mathrm dy}{\mathrm dx} = \frac{\frac{\mathrm d}{\mathrm dx}(\sqrt x-1)\cdot(\sqrt x+1)-(\sqrt x-1)\cdot \frac{\mathrm d}{\mathrm dx}(\sqrt x+1)}{(\sqrt x+1)^2}\]

It might me easiest to calculate \[f'=\frac{\mathrm d}{\mathrm dx}(\sqrt x-1)\]
and\[g'=\frac{\mathrm d}{\mathrm dx}(\sqrt x+1)\]first

Answer 14

\[f'=\frac{\mathrm d}{\mathrm dx}(\sqrt x-1)\\\quad=\frac{\mathrm d}{\mathrm dx}(x^{1/2}-x^0)\\\quad=\tfrac12x^{1/2-1}-0\, x^{0-1}\\\quad=\tfrac12x^{-1/2}\\\quad=\frac1{2\sqrt x}\]

Answer 15

thank you @UnkleRhaukus for your help