Chapter 3.1, Problem 23E

To determine

To find:

The differentiation of the given function

Answer to Problem 23E

Differentiating the function is $\frac{dy}{dt}=\frac{1}{5}{t}^{-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+10{\left(t\right)}^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{t}^4$

Explanation of Solution

Given:

The function is

  $f\left(t\right)=\sqrt[5]{t}+4\sqrt{t^5}$

Concept used:

Definition of the differentiation:-Differentiation is the action of computing a derivative

The derivative of a function $y=f\left(x\right)$ of a variable $x$ is a measure of the rate at which the value y of the function changes with respect to $x$

Calculation:

The function

  $f\left(t\right)=\sqrt[5]{t}+4\sqrt{t^5}\mathrm{...................}\left(1\right)$

The derivative of a function

  $\begin{array}{l}y=f\left(x\right)\\ y^{\prime }=f^{\prime }\left(x\right)=\frac{dy}{dx}\end{array}$

Differentiating the equation (1) with respect to $t$

  $\begin{array}{l}\frac{dy}{dt}=\frac{d}{dt}\left(\sqrt[5]{t}+4\sqrt{t^5}\right)\\ \frac{dy}{dt}=\frac{d}{dt}\left(\sqrt[5]{t}\right)+4\frac{d}{dt}\sqrt{t^5}\\ \frac{dy}{dt}=\frac{1}{5}{t}^{-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+\frac{4}{2}{\left(t^5\right)}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}5t^4\\ \frac{dy}{dt}=\frac{1}{5}{t}^{-\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}+10{\left(t\right)}^{\raisebox{1ex}{$-5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{t}^4\end{array}$