Logarthimic differentiation

lamaclass

Junior Member
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Oct 18, 2009
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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)]=?
 
Hello, lamaclass!

You were doing fine . . .


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

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

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

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


 
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