Question

Differentiate the function.

10. \(g\left( t \right) = \sqrt {1 + \ln t} \)

(g\left( t \right) = \sqrt {1 + \ln t}

Short answer

The differentiation of\(g\left( t \right) = \sqrt {1 + \ln t} \) is \(\frac{1}{{2t\sqrt {1 + \ln t} }}\).

Step-by-step solution

Formula used 

1. Differential formula:\(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\)
2. Power rule: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)
3. \(\frac{d}{{dx}}\left( {{b^n}} \right) = 0\), where \(b,n\) are constants

Differentiate the given function

Differentiate\(g\left( t \right) = \sqrt {1 + \ln t} \)with respect to \(t\) by using\(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\).

\(\begin{aligned}{c}g'\left( t \right)&= \frac{d}{{dt}}\left( {\sqrt {1 + \ln t} } \right)\\&= \frac{d}{{dt}}{\left( {1 + \ln t} \right)^{\frac{1}{2}}}\\&= \frac{1}{2}{\left( {1 + \ln t} \right)^{\frac{1}{2} - 1}} \cdot \frac{d}{{dt}}\left( {1 + \ln t} \right)\\&= \frac{1}{2}{\left( {1 + \ln t} \right)^{ - \frac{1}{2}}} \cdot \left( {\frac{d}{{dt}}\left( 1 \right) + \frac{d}{{dt}}\left( {\ln t} \right)} \right)\end{aligned}\)

Simplify further by using \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) and \(\frac{d}{{dx}}\left( {{b^n}} \right) = 0\).

\(\begin{aligned}{c}g'\left( t \right)&= \frac{1}{2}{\left( {1 + \ln t} \right)^{ - \frac{1}{2}}} \cdot \left( {0 + \frac{1}{t}} \right)\\&= \frac{1}{{2t\sqrt {1 + \ln t} }}\end{aligned}\)

Hence, the differentiation of \(g\left( t \right) = \sqrt {1 + \ln t} \) is \(\frac{1}{{2t\sqrt {1 + \ln t} }}\).