# Find the first differential of y= Ln [( x^(1/2)) / ( 1+x^(1/2))] ?

Jun 19, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(x + {x}^{\frac{3}{2}}\right)}$

#### Explanation:

I have asumed that the question as written has omitted brackets around the $\left(\frac{1}{2}\right)$ exponents so that:
$y = L n \left[\frac{{x}^{\frac{1}{2}}}{1 + {x}^{\frac{1}{2}}}\right]$

If this is the case then:

dy/dx = (1+x^(1/2))/(x^(1/2) * $\frac{d}{\mathrm{dx}} \left(\frac{{x}^{\frac{1}{2}}}{1 + {x}^{\frac{1}{2}}}\right)$ Standard differential and chain rule

dy/dx = (1+x^(1/2))/(x^(1/2) * $\frac{\left(1 + {x}^{\frac{1}{2}}\right) \cdot \frac{1}{2} {x}^{- \frac{1}{2}} - {x}^{\frac{1}{2}} \cdot \frac{1}{2} {x}^{- \frac{1}{2}}}{1 + {x}^{\frac{1}{2}}} ^ 2$ Quotient rule

= (1+x^(1/2))/(x^(1/2)  * $\frac{1}{2} {x}^{- \frac{1}{2}} \frac{\left(1 + {x}^{\frac{1}{2}} - {x}^{\frac{1}{2}}\right)}{1 + {x}^{\frac{1}{2}}} ^ 2$

= (1+x^(1/2))/(x^(1/2)  * $\frac{\left(1 + \cancel{{x}^{\frac{1}{2}}} - \cancel{{x}^{\frac{1}{2}}}\right)}{2 {x}^{\frac{1}{2}} {\left(1 + {x}^{\frac{1}{2}}\right)}^{2}}$

$= \frac{\left(1 + {x}^{\frac{1}{2}}\right)}{2 x {\left(1 + {x}^{\frac{1}{2}}\right)}^{2}}$

$= \frac{\cancel{\left(1 + {x}^{\frac{1}{2}}\right)}}{2 x {\left(1 + {x}^{\frac{1}{2}}\right)}^{\cancel{2}}}$

$= \frac{1}{2 \left(x + {x}^{\frac{3}{2}}\right)}$