Aus Online Mathematik Brückenkurs 2

There is a "ln of something", so a first step in the differentiation is to take the derivative of the logarithm:

\displaystyle \frac{d}{dx}\ln \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)=\frac{1}{\left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right.}\centerdot \left( \left\{ \left. \sqrt{x}+\sqrt{x+1} \right\} \right. \right)^{\prime }

We can carry out the differentiation of \displaystyle \sqrt{x}+\sqrt{x+1} on the right-hand side term by term to obtain

\displaystyle \quad =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \left( \sqrt{x} \right)^{\prime }+\left( \sqrt{x+1} \right)^{\prime } \right]

and it remains then only to differentiate \displaystyle \sqrt{x},which we do directly, and \displaystyle \sqrt{x+1} which a simple inner derivative)

\displaystyle

\displaystyle \begin{align} & =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot \left( x+1 \right)^{\prime } \right] \\ & =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{x+1}}\centerdot 1 \right] \\ \end{align}

If we rewrite the expression inside the square brackets using a common denominator, we get

\displaystyle =\frac{1}{\sqrt{x}+\sqrt{x+1}}\centerdot \left[ \frac{\sqrt{x+1}+\sqrt{x}}{2\sqrt{x}\sqrt{x+1}} \right],

and we can then eliminate the factor \displaystyle \sqrt{x+1}+\sqrt{x} from the numerator and denominator to get

\displaystyle =\frac{1}{2\sqrt{x}\sqrt{x+1}}