text: # What is the derivative of arctan(1/x)?
  
  Jul 31, 2015
  
  The derivative is: $\frac{- 1}{{x}^{2} + 1}$
  
  #### Explanation:
  
  $\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{1 + {x}^{2}}$
  
  So
  
  $\frac{d}{\mathrm{dx}} \arctan \left(u\right) = \frac{1}{1 + {u}^{2}} \frac{\mathrm{du}}{\mathrm{dx}}$
  
  And
  
  $\frac{d}{\mathrm{dx}} \arctan \left(\frac{1}{x}\right) = \frac{1}{1 + {\left(\frac{1}{x}\right)}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$
  
  $= \frac{1}{1 + \frac{1}{x} ^ 2} \cdot \frac{- 1}{x} ^ 2$
  
  $= {x}^{2} / \left({x}^{2} + 1\right) \cdot \frac{- 1}{x} ^ 2$
  
  $= \frac{- 1}{{x}^{2} + 1}$
  
  Faster Method?
  
  Use the fact that $\arctan \left(\frac{1}{x}\right) = a r c \cot \left(x\right)$
  
  and
  
  $\frac{d}{\mathrm{dx}} a r c \cot \left(x\right) = - \frac{1}{1 + {x}^{2}}$
  
  to go straight to the answer.
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