- #1
TommG
- 28
- 0
Need to find derivative using logarithmic differentiation
[itex]y = \sqrt{x(x+1)}[/itex]
My attempt
[itex]ln y = ln \sqrt{x(x+1)}[/itex]
[itex]ln y = \frac{1}{2}ln x(x+1)[/itex]
[itex]ln y = \frac{1}{2}ln x + ln(x+1)[/itex]
[itex]\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}[/itex]
[itex]\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}[/itex]
[itex]\frac{dy}{dx}= y(\frac{1}{2x} + \frac{1}{x+1})[/itex]
[itex]\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{1}{2x} + \frac{1}{x+1})[/itex]
[itex]\frac{dy}{dx}= \frac{\sqrt{x(x+1)}}{2x} + \frac{\sqrt{x(x+1)}}{x+1}[/itex]
answer in book [itex]\frac{2x+1}{2\sqrt{x(x+1))}} [/itex]
[itex]y = \sqrt{x(x+1)}[/itex]
My attempt
[itex]ln y = ln \sqrt{x(x+1)}[/itex]
[itex]ln y = \frac{1}{2}ln x(x+1)[/itex]
[itex]ln y = \frac{1}{2}ln x + ln(x+1)[/itex]
[itex]\frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1}[/itex]
[itex]\frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1}[/itex]
[itex]\frac{dy}{dx}= y(\frac{1}{2x} + \frac{1}{x+1})[/itex]
[itex]\frac{dy}{dx}= \sqrt{x(x+1)}(\frac{1}{2x} + \frac{1}{x+1})[/itex]
[itex]\frac{dy}{dx}= \frac{\sqrt{x(x+1)}}{2x} + \frac{\sqrt{x(x+1)}}{x+1}[/itex]
answer in book [itex]\frac{2x+1}{2\sqrt{x(x+1))}} [/itex]