Question

3-34 Differentiate the function.

13. \(s\left( t \right) = \frac{1}{t} + \frac{1}{{{t^2}}}\)

Short answer

The differentiation of the function \(s\left( t \right) = \frac{1}{t} + \frac{1}{{{t^2}}}\) is \(s'\left( t \right) = - \frac{1}{{{t^2}}} - \frac{2}{{{t^3}}}\).

Step-by-step solution

Write the formula of the sum rules and the power rule

The Sum Rule:\(\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\)

The Power Rule: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)where \(n\) is a positive integer

Find the differentiation of the function

Consider the function \(s\left( t \right) = \frac{1}{t} + \frac{1}{{{t^2}}}\). Differentiate the function w.r.t \(t\) by using the sum rule and the power rule.

\(\begin{aligned}\frac{{d\left( {s\left( t \right)} \right)}}{{dt}} &= \frac{{d\left( {\frac{1}{t} + \frac{1}{{{t^2}}}} \right)}}{{dt}}\\ &= \frac{d}{{dt}}\left( {\frac{1}{t}} \right) + \frac{d}{{dt}}\left( {\frac{1}{{{t^2}}}} \right)\\ &= \frac{d}{{dt}}\left( {{t^{ - 1}}} \right) + \frac{d}{{dt}}\left( {{t^{ - 2}}} \right)\\ &= \frac{d}{{dt}}\left( { - 1 \cdot {t^{ - 1 - 1}}} \right) + \frac{d}{{dt}}\left( { - 2 \cdot {t^{ - 2 - 1}}} \right)\\ &= - {t^{ - 2}} - 2{t^{ - 3}}\\ &= - \frac{1}{{{t^2}}} - \frac{2}{{{t^3}}}\end{aligned}\)

Thus, the derivative of the function \(s\left( t \right) = \frac{1}{t} + \frac{1}{{{t^2}}}\) is \(s'\left( t \right) = - \frac{1}{{{t^2}}} - \frac{2}{{{t^3}}}\).