Question

3-34 Differentiate the function.

8. \(f\left( t \right) = {t^3} + {e^3}\)

Short answer

The differentiation of the function \(f\left( t \right) = {t^3} + {e^3}\) is \(f'\left( t \right) = 3{t^2}\).

Step-by-step solution

Write the formula of the sum rule, derivative of a constant function and the power rule

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

Derivative of a Constant Function:\(\frac{d}{{dx}}\left( c \right) = 0\)

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 \(f\left( t \right) = {t^3} + {e^3}\). Differentiate the function w.r.t \(t\) by using the derivative of a constant function and the power rule.

\(\begin{aligned}\frac{{d\left( {f\left( t \right)} \right)}}{{dt}} &= \frac{{d\left( {{t^3} + {e^3}} \right)}}{{dt}}\\ &= \frac{d}{{dt}}\left( {{t^3}} \right) + \frac{d}{{dt}}\left( {{e^3}} \right)\\ &= \frac{d}{{dt}}\left( {3{t^{3 - 1}}} \right) + 0\\ &= 3{t^2}\end{aligned}\)

Thus, the derivative of the function \(f\left( t \right) = {t^3} + {e^3}\) is \(f'\left( t \right) = 3{t^2}\).