# How do you differentiate f(x)=(1+arctanx)/(2-3arctanx)?

Mar 24, 2017

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{5}{\left(1 + {x}^{2}\right) {\left(2 - 3 \arctan x\right)}^{2}}$

#### Explanation:

Here we can use quotient rule.

Also note that derivative of $\arctan x$, which is also known as ${\tan}^{- 1} x$ is $\frac{1}{1 + {x}^{2}}$ i.e.

$\frac{d}{\mathrm{dx}} \arctan x = \frac{1}{1 + {x}^{2}}$

Now according to quotient rule if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$

then $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{h \left(x\right) \cdot \frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} - g \left(x\right) \cdot \frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}}}{{\left(h \left(x\right)\right)}^{2}}$

Here $g \left(x\right) = 1 + \arctan x$ hence $\frac{\mathrm{dg} \left(x\right)}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$

and $h \left(x\right) = 2 - 3 \arctan x$ hence $\frac{\mathrm{dh} \left(x\right)}{\mathrm{dx}} = - \frac{3}{1 + {x}^{2}}$

Hence

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\left(2 - 3 \arctan x\right) \cdot \frac{1}{1 + {x}^{2}} - \left(1 + \arctan x\right) \left(- \frac{3}{1 + {x}^{2}}\right)}{{\left(2 - 3 \arctan x\right)}^{2}}$

= $\frac{\left(\frac{2}{1 + {x}^{2}} - \frac{3 \arctan x}{1 + {x}^{2}} + \frac{3}{1 + {x}^{2}} + \frac{3 \arctan x}{1 + {x}^{2}}\right)}{{\left(2 - 3 \arctan x\right)}^{2}}$

= $\frac{\frac{5}{1 + {x}^{2}}}{{\left(2 - 3 \arctan x\right)}^{2}}$

= $\frac{5}{\left(1 + {x}^{2}\right) {\left(2 - 3 \arctan x\right)}^{2}}$