Sunday

March 29, 2015

March 29, 2015

Posted by **Amy** on Monday, June 29, 2009 at 8:21pm.

Using my notes, I came up with f'(x) = (1/2*square root of x)/x

The answer in the back of the book says f'(x) = (2+ln x)/(2*square root of x).

Can someone please explain how? Thanks.

- Calculus/Logarithms -
**Amy**, Monday, June 29, 2009 at 8:32pmI think I figured it out. I believe it's just a matter of applying the Chain Rule.

- Calculus/Logarithms -
**Damon**, Monday, June 29, 2009 at 8:37pmy = (x^.5) ln x

dy/dx = (x^.5) /x + ln x (.5 x^-.5)

but x^.5 / x = x^-.5 so

= (x^-.5) (1+.5 ln x)

= (1+.5 ln x) / x^.5

= (2+1 ln x)/2 x^.5

so I agree with the book.

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