Easy derivitive giving me some trouble

  • Thread starter kdinser
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  • #1
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Ok, this one is giving me a headache. I'm missing something stupid here and I don't know where, probably because I've been studying since 4:30 am EST :smile:

find d/dx (ln x^(1/2) I know the answer is 1/2x as it says in the back of the book, I've also confirmed that the answer is correct with a graphing calculator

I use the chain rule, f'(g(x))(g'(x)) if f(x)=ln x and g(x)=x^(1/2) then I have [1/x^(1/2)][1/(2x(x^(1/2))]

Now it looks to me like the end of this problem would be 1/2x^2 Since this is not the correct derivative, something is wrong somewhere. I'm sure it's a silly algebraic mistake where x^(1/2) should be in the numerator at some point to allow it to cancel out. thanks for any help.

Also, if these forums have some kind of standard that problems and work should be submitted in, please point me to it. I'd hate to think I'm making things more difficult with the way I'm formatting my problems.
 

Answers and Replies

  • #2
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ln x^1/2

because of the rules of ln, you can 'pull' the 1/2 down and out in front

1/2 ln x

and as you know, the deriv. of ln x, is 1/x

1/2 1/x

result

1/2x

;)
 
  • #3
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DOH!!! I knew I was forgetting something simple, just trying to crank through these exercises to fast I guess, thanks.
 
  • #4
Doc Al
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kdinser said:
I use the chain rule, f'(g(x))(g'(x)) if f(x)=ln x and g(x)=x^(1/2) then I have [1/x^(1/2)][1/(2x(x^(1/2))]
f'(g(x)) = 1/x^(1/2) = x^(-1/2)
g'(x) = (1/2)x^(-1/2)

Now the derivative becomes:
x^(-1/2)(1/2)x^(-1/2) = 1/2 x^(-1) = 1/(2x)
 
Last edited:

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