Question

In Exercises \(13-22,\) find the derivatives of \(y\) with respect to the appropriate variable. $$y=\cot ^{-1} \frac{1}{x}-\tan ^{-1} x$$

Short answer

The derivative of the function \(y=\cot ^{-1} \frac{1}{x}-\tan ^{-1} x\) is \(0\).

Step-by-step solution

Identify the Function

The function that we need to differentiate is \(y=\cot ^{-1} \frac{1}{x}-\tan ^{-1} x\).

Application of Chain Rule - Part 1

First, differentiate \(\cot^{-1}(1/x)\). The derivative of \(\cot^{-1}(u)\) with respect to \(u\) is \(-1/(1+u^2)\). By applying chain rule \(dy/du = dy/du \cdot du/dx\), where \(u = 1/x\), the derivative of \(1/x\) with respect to \(x\) is \(-1/x^2\). Therefore: \[ \frac{d(\cot^{-1}(1/x))}{dx} = - \frac{1}{1+(1/x)^2} \cdot -1/x^2 = \frac{1}{1+x^2} \cdot 1= \frac{1}{1+x^2}\]

Application of Chain Rule - Part 2

Next, differentiate \(-\tan^{-1}(x)\). The derivative of \(-\tan^{-1}(x)\) with respect to \(x\) is \(-(1/(1+x^2))\). Therefore: \[\frac{d( - \tan^{-1}(x))}{dx} = - \frac{1}{1+x^2}\]

Find the Derivative

The derivative of function \(y\) with respect to \(x\) (denoted \(y'\)) is obtained by adding the derivatives obtained in the previous steps: \[y' = \frac{1}{x^2+1} - \frac{1}{x^2+1} = 0.\] Hence \(y'\) is \(0\).