How do you find the derivative of $ln \sqrt{x}$ $lnsqrt x$ ?

Question

How do you find the derivative of $ln \sqrt{x}$ $lnsqrt x$ ?

1 Answer

Impact of this question

1686 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-09T20:53:08

In general
$\frac{d ln (a)}{d a} = \frac{1}{a}$ $(d ln(a))/(da) = 1/a$

Specifically if $a = \sqrt{x}$ $a=sqrt(x)$
$\frac{d ln (\sqrt{x})}{d \sqrt{x}} = \frac{1}{\sqrt{x}}$ $(d ln(sqrt(x)))/(d sqrt(x)) = 1/sqrt(x)$

$\sqrt{x} = x^{\frac{1}{2}}$ $sqrt(x) = x^(1/2)$

$\frac{d \sqrt{x}}{d x} = \frac{1}{2} x^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$ $(d sqrt(x))/dx = 1/2 x^(-1/2) = 1/(2sqrt(x))$

$\frac{d ln (\sqrt{x})}{d x} = \frac{d ln (\sqrt{x})}{d \sqrt{x}} \cdot \frac{d \sqrt{x}}{d x}$ $(d ln(sqrt(x)))/(dx) = (d ln(sqrt(x)))/(d sqrt(x)) * (d sqrt(x))/dx$

$= \frac{1}{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ $= 1/sqrt(x) * 1/(2sqrt(x))$

$= \frac{1}{2 x}$ $= 1/(2x)$

That is, the derivative of $ln \sqrt{x}$ $ln sqrt(x)$
is
$\frac{1}{2 x}$ $1/(2x)$

Science

Math

Humanities

... and beyond

How do you find the derivative of $ln \sqrt{x}$ $lnsqrt x$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the derivative of lnsqrt xln√xlnsqrt x?

1 Answer

Related questions

Impact of this question

How do you find the derivative of $ln \sqrt{x}$ $lnsqrt x$ ?