Chapter 2, Problem 28RE

To determine

To find: The derivative of the function.

Answer to Problem 28RE

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

Explanation of Solution

Given information:

A function, $y=\frac{t^3}{{\pi }^2}-\frac{{\pi }^3}{t^2}$ .

Concept Used

  $\frac{d\left(au\left(x\right)+bv\left(x\right)\right)}{dx}=a\frac{d\left(u\right)}{dx}+b\frac{d\left(v\right)}{dx}$

where $a$ and $b$ are constants $u\left(x\right)$ and $v\left(x\right)$ are functions. of $x$ .

Calculation:

Differentiate $y=\frac{t^3}{{\pi }^2}-\frac{{\pi }^3}{t^2}$ as follows:

  $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left[\frac{t^3}{{\pi }^2}-\frac{{\pi }^3}{t^2}\right]\\ =\frac{1}{{\pi }^2}\cdot \frac{d}{dt}\left[t^3\right]-{\pi }^3\cdot \frac{d}{dt}\left[\frac{1}{t^2}\right]\\ =\frac{1}{{\pi }^2}\cdot 3t^2-{\pi }^3\cdot \left(-2t^{-3}\right)\\ =\frac{3t^2}{{\pi }^2}+\frac{2{\pi }^3}{t^3}\end{array}$

Conclusion:

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