How do you find the derivative of $ln x^(1/2)$ ?

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Answer 1 · 2015-05-14T12:37:14

Remembering the rule to derive $ln$ functions:

If $y = ln(f(x))$ , $(dy)/(dx) = (f'(x))/(f(x))$ .

And using the chain rule, we can say that $u=x^(1/2)$

Now, $(dy)/(du) = (u')/(u)$

$u' = (1/2)*x^(-1/2) = 1/(2x^(1/2))$

Substituting in our original derivative:

$(dy)/(dx) = (1/(2x^(1/2)))/x^(1/2)$

$(dy)/(dx) = 1/(2x^(1/2))*1/x^(1/2) = 1/(2x)$

Answer 2 · 2015-05-14T13:12:18

There are a couple of methods available. Here is one:

Use properties of $ln$ to rewrite:

$lnx^(1/2) =1/2 lnx$

So the derivative is

$1/2*1/x =1/(2x)$

How do you find the derivative of ln x^(1/2)ln x^(1/2)?