Chapter 3, Problem 27RE

To determine

To find:The derivative of the given function.

Answer to Problem 27RE

The derivative function of $y=\frac{t^2}{{\pi }^3}-\frac{{\pi }^2}{t^3}$ is $\frac{dy}{dt}=\frac{2t}{{\pi }^3}+\frac{3{\pi }^2}{t^4}$ .

Explanation of Solution

Given information: $y=\frac{t^2}{{\pi }^3}-\frac{{\pi }^2}{t^3}$ .

Formula used:

• Derivative rule: $\left(x^n\right)\prime =n\left(x^{n-1}\right)$

Calculation:

  $\begin{array}{l}y=\frac{t^2}{{\pi }^3}-\frac{{\pi }^2}{t^3}\\ y=\frac{1}{{\pi }^3}{t}^2-{\pi }^2{t}^{-3}\\ \frac{dy}{dt}=\frac{d}{dt}\left(\frac{1}{{\pi }^3}{t}^2-{\pi }^2{t}^{-3}\right)\\ \frac{dy}{dt}=\frac{1}{{\pi }^3}\frac{d}{dt}{t}^2-{\pi }^2\frac{d}{dt}{t}^{-3}\\ \frac{dy}{dt}=\frac{1}{{\pi }^3}\left(2t\right)-{\pi }^2\left(3t^{-4}\right)\\ \frac{dy}{dt}=\frac{2t}{{\pi }^3}-\frac{3{\pi }^2}{t^4}\end{array}$

The derivative function of $y=\frac{t^2}{{\pi }^3}-\frac{{\pi }^2}{t^3}$ is $\frac{dy}{dt}=\frac{2t}{{\pi }^3}+\frac{3{\pi }^2}{t^4}$ .