Find the derivative of f(x) = (square root of x)(ln x).
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.
2 answers
I think I figured it out. I believe it's just a matter of applying the Chain Rule.
y = (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.
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.
0 answers