Chapter A6, Problem 16E

To determine

To find: the derivative of the given equation of y with respect to appropriate variable.

Answer to Problem 16E

The derivative is $\frac{dy}{dt}=-sec{h}^2(\frac{1}{t})+2t\tanh (\frac{1}{t})$ .

Explanation of Solution

Given:

  $y=t^2\tanh (\frac{1}{t})$

Calculation:

The derivative of y with respect to t is calculated below

  $y=t^2\tanh (\frac{1}{t})$

Differentiating y with respect to t

  $\begin{array}{l}\frac{dy}{dt}=[sec{h}^2(t^{-1})](-t^{-2})(t^2)+2t\tanh (\frac{1}{t})\\ \frac{dy}{dt}=-sec{h}^2(\frac{1}{t})+2t\tanh (\frac{1}{t})\end{array}$

Conclusion:

Therefore, the derivative is $\frac{dy}{dt}=-sec{h}^2(\frac{1}{t})+2t\tanh (\frac{1}{t})$ .