# What is the derivative of arctan(13/x) - arctan(3/x)?

Jun 2, 2015

By derivate rules definition, we have that

Be $y = \arctan u$, then $y ' = \frac{u '}{1 + {u}^{2}}$

Thus, in this case let's use chain rule, which states

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$, and rename $u = \frac{13}{x} = 13 {x}^{-} 1$, and, alike, $v = \frac{3}{x} = 3 {x}^{-} 1$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 13 {x}^{-} 2 \frac{- 13 {x}^{-} 2}{1 + {\left(\frac{13}{x}\right)}^{2}} - \left(- 3 {x}^{-} 2 \frac{- 3 {x}^{-} 2}{1 + {\left(\frac{3}{x}\right)}^{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{13}{x} ^ 2 \frac{- 13 \cancel{{x}^{2}}}{\cancel{{x}^{2}} \left({x}^{2} + 169\right)} + \frac{3}{x} ^ 2 \frac{3 \cancel{{x}^{2}}}{\cancel{{x}^{2}} \left({x}^{2} + 9\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{169}{{x}^{2} \left({x}^{2} + 169\right)} + \frac{9}{{x}^{2} \left({x}^{2} + 9\right)}$

Thus, the l.c.d. is ${x}^{2} \left({x}^{2} + 169\right) \left({x}^{2} + 9\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{169 \left({x}^{2} \left({x}^{2} + 9\right)\right) + 9 \left({x}^{2} \left({x}^{2} + 169\right)\right)}{{x}^{2} \left({x}^{2} + 169\right) \left({x}^{2} + 9\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{169 {x}^{4} + 1521 {x}^{2} + 9 {x}^{4} + 1521 {x}^{2}}{{x}^{2} \left({x}^{2} + 169\right) \left({x}^{2} + 9\right)}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{178 {x}^{4} + 3042 {x}^{2}}{{x}^{2} \left({x}^{2} + 169\right) \left({x}^{2} + 9\right)} = \frac{178 {x}^{2} + 3042}{\left({x}^{2} + 169\right) \left({x}^{2} + 9\right)}$