Question

3-34 Differentiate the function.

14. \(r\left( t \right) = \frac{a}{{{t^2}}} + \frac{b}{{{t^4}}}\)

Short answer

The differentiation of the function \(r\left( t \right) = \frac{a}{{{t^2}}} + \frac{b}{{{t^4}}}\) is \(r'\left( t \right) = - \frac{{2a}}{{{t^3}}} - \frac{{4b}}{{{t^5}}}\).

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 \(r\left( t \right) = \frac{a}{{{t^2}}} + \frac{b}{{{t^4}}}\). Differentiate the function w.r.t \(t\) by using the sum rule and the power rule.

\(\begin{aligned}\frac{{d\left( {r\left( t \right)} \right)}}{{dt}} &= \frac{{d\left( {\frac{a}{{{t^2}}} + \frac{b}{{{t^4}}}} \right)}}{{dt}}\\ &= \frac{d}{{dt}}\left( {\frac{a}{{{t^2}}}} \right) + \frac{d}{{dt}}\left( {\frac{b}{{{t^4}}}} \right)\\ &= \frac{d}{{dt}}\left( {a{t^{ - 2}}} \right) + \frac{d}{{dt}}\left( {b{t^{ - 4}}} \right)\\ &= a\frac{d}{{dt}}\left( { - 2 \cdot {t^{ - 2 - 1}}} \right) + b\frac{d}{{dt}}\left( { - 4 \cdot {t^{ - 4 - 1}}} \right)\\ &= - 2a{t^{ - 3}} - 4b{t^{ - 5}}\\ &= - \frac{{2a}}{{{t^3}}} - \frac{{4b}}{{{t^5}}}\end{aligned}\)

Thus, the derivative of the function \(r\left( t \right) = \frac{a}{{{t^2}}} + \frac{b}{{{t^4}}}\) is \(r'\left( t \right) = - \frac{{2a}}{{{t^3}}} - \frac{{4b}}{{{t^5}}}\).