Chapter 4, Problem 25RE

To determine

To Find: The derivative of the given function.

Answer to Problem 25RE

The derivative is $\frac{1}{\sqrt{t^2-1}}+{\sec }^{-1}t-\frac{1}{2t}$ .

Explanation of Solution

Given information: Given function $y=t{\sec }^{-1}t-\frac{1}{2}\ln t$

Formula used: $\frac{d(\ln x)}{dx}=\frac{1}{x},\frac{d(x)}{dx}=1\text{and}\frac{d({\sec }^{-1}x)}{dx}=\frac{1}{x\sqrt{x^2-1}}$ with product and subtraction rule..

Calculation:

  $\begin{array}{l}y=t{\sec }^{-1}t-\frac{1}{2}\ln t\\ \text{Differentiate}\text{with}\text{respect}\text{to}t\\ \frac{dy}{dt}=\left(t\times \frac{d({\sec }^{-1}t)}{dt}+\frac{d(t)}{dt}\times {\sec }^{-1}t\right)-\frac{1}{2}\frac{d(\ln t)}{dt}\\ \frac{dy}{dt}=t\times \frac{1}{t\sqrt{t^2-1}}+1\times {\sec }^{-1}t-\frac{1}{2t}\\ \frac{dy}{dt}=\frac{1}{\sqrt{t^2-1}}+{\sec }^{-1}t-\frac{1}{2t}\end{array}$