Logarthimic differentiation

[lamaclass]

I am having a hard time with finishing this problem. Can anyone show what to do for the step that I'm stuck on? Thanks!

y=x(x[sup:3uveu773]2[/sup:3uveu773]-1)[sup:3uveu773]1/2[/sup:3uveu773]

ln y=ln x+1/2 ln(x[sup:3uveu773]2[/sup:3uveu773]+1)

1/y*dy/dx=1/x+1/2*(1/x[sup:3uveu773]2[/sup:3uveu773]-1)

1/y*dy/dx=2x[sup:3uveu773]2[/sup:3uveu773]-1/x(x[sup:3uveu773]2[/sup:3uveu773]-1)

dy/dx=y[2x[sup:3uveu773]2[/sup:3uveu773]-1/x(x[sup:3uveu773]2[/sup:3uveu773]-1)]=?

 

[soroban]

Hello, lamaclass!

You were doing fine . . .

$\displaystyle y\;=\;x\left(x^2-1\right)^{\frac{1}{2}}$

$\displaystyle \ln y \;=\;\ln x + \tfrac{1}{2}\ln(x^2-1)$

$\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\frac{1}{x} + \frac{1}{2}\,\frac{2x}{x^2+1} \;=\;\frac{1}{x} + \frac{x}{x^2-1} \;=\;\frac{2x^2-1}{x(x^2-1)}$

$\displaystyle \frac{dy}{dx} \;=\;y\left[\frac{2x^2-1}{x(x^2+1)}\right] \;=\;x(x^2-1)^{\frac{1}{2}}\cdot\frac{2x^2-1}{x(x^2-1)} \;=\; \frac{2x^2-1}{(x^2-1)^{\frac{1}{2}}}$