University Calculus: Early Transcendentals (3rd Edition)

Given that $y$ = $\cot^{-1}{\frac{{1}}{x}}-\tan^{-1}{x}$, on applying the product rule of the derivative, we get: $\frac{dy}{dx}$ = ${-\frac{1}{1+{(\frac{1}{x})^2}}\frac{d(\frac{1}{x})}{dx}-\frac{1}{{x^2+1}}}$ or $\frac{dy}{dx}$ = ${ \frac {x^2}{x^2({x^2+1})}-{\frac{1}{{(x^2+1)}}}}$ = 0 The final answer is: = 0