Question

3-34 Differentiate the function.

4. \(g\left( t \right) = 5t + 4{t^2}\)

Short answer

The differentiation of the function \(g\left( t \right) = 5t + 4{t^2}\) is \(g'\left( t \right) = 5 + 8t\).

Step-by-step solution

Write the formula of the sum rule, the power function 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 function: \(\frac{d}{{dx}}\left( x \right) = 1\)

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 \(g\left( t \right) = 5t + 4{t^2}\). Differentiate the function w.r.t \(t\) by using the sum rule, the power function and the power rule.

\(\begin{aligned}\frac{{d\left( {g\left( t \right)} \right)}}{{dt}} &= \frac{{d\left( {5t + 4{t^2}} \right)}}{{dt}}\\ &= \frac{d}{{dt}}\left( {5t} \right) + \frac{d}{{dt}}\left( {4{t^2}} \right)\\ &= 5\frac{d}{{dt}}\left( t \right) + 4\frac{d}{{dt}}\left( {{t^2}} \right)\\ &= 5\left( 1 \right) + 4\left( {2{t^{2 - 1}}} \right)\\ &= 5 + 8t\end{aligned}\)

Thus, the derivative of the function \(g\left( t \right) = 5t + 4{t^2}\) is \(g'\left( t \right) = 5 + 8t\).