Chapter 3, Problem 28RE

To determine

To calculate: The derivative of the function.

Answer to Problem 28RE

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

Explanation of Solution

Given Information: The function is $y=\frac{t^3}{{\pi }^2}-\frac{{\pi }^3}{t^2}$ .

Calculation:

The derivative of the function y with respect to t is calculated as,

  $\begin{array}{c}y=\frac{t^3}{{\pi }^2}-\frac{{\pi }^3}{t^2}\\ \frac{dy}{dt}=\frac{d}{dt}\left(\frac{t^3}{{\pi }^2}-\frac{{\pi }^3}{t^2}\right)\left(\text{Substitute the value of }y\text{ in }\frac{dy}{dt}\right)\\ =\frac{d}{dt}\left(\frac{t^3}{{\pi }^2}\right)-\frac{d}{dt}\left(\frac{{\pi }^3}{t^2}\right)\left(\text{Use Distributive property}\right)\\ =\frac{1}{{\pi }^2}\frac{d}{dt}\left(t^3\right)-{\pi }^3\frac{d}{dt}\left(t^{-2}\right)\left(\text{Take the constant term outside the derivative}\right)\\ =\frac{1}{{\pi }^2}\left(3t^2\right)-{\pi }^3\left(-2t^{-2-1}\right)\left(\frac{d}{dt}\left(t^m\right)=m{t}^{m-1}\right)\\ =\frac{3t^2}{{\pi }^2}-{\pi }^3\left(-2t^{-3}\right)\left(\text{Simplify the expression}\right)\\ =\frac{3t^2}{{\pi }^2}+2{\pi }^3{t}^{-3}\\ =\frac{3t^2}{{\pi }^2}+\frac{{\pi }^3}{t^3}\end{array}$

Conclusion:

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