Question

Differentiate the function.

12.\(p\left( t \right) = \ln \sqrt {{t^2} + 1} \).

Short answer

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

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 combined with the chain rule: \(\frac{d}{{dx}}\left( {{u^n}} \right) = n{u^{n - 1}}\frac{{du}}{{dx}}\)

Differentiate the given function

Differentiate\(p\left( t \right) = \ln \sqrt {{t^2} + 1} \)with respect to \(t\) by using\(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\).

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

Simplify further by using power rule combined with the chain rule.

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

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