# How do you differentiate y=sinx+x^2tan^-1x?

Jun 13, 2018

$y ' \left(x\right) = \cos \left(x\right) + 2 x \arctan \left(x\right) + {x}^{2} / \left(1 + {x}^{2}\right)$

#### Explanation:

Note that
$\left(\arctan \left(x\right)\right) ' = \frac{1}{1 + {x}^{2}}$
and by the sum and product rule we get

$y ' \left(x\right) = \cos \left(x\right) + 2 x \arctan \left(x\right) + {x}^{2} / \left(1 + {x}^{2}\right)$

Jun 13, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x + {x}^{2} / \left(1 + {x}^{2}\right) + 2 x {\tan}^{-} 1 x$

#### Explanation:

$\text{differentiate "x^2tan^-1x" using the "color(blue)"product rule}$

$\text{given "y=f(x)g(x)" then}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x\right) g ' \left(x\right) + g \left(x\right) f ' \left(x\right) \leftarrow \textcolor{b l u e}{\text{product rule}}$

$f \left(x\right) = {x}^{2} \Rightarrow f ' \left(x\right) = 2 x$

$g \left(x\right) = {\tan}^{-} 1 x \Rightarrow g ' \left(x\right) = \frac{1}{1 + {x}^{2}}$

$\frac{d}{\mathrm{dx}} \left({x}^{2} {\tan}^{-} 1 x\right)$

$= {x}^{2} / \left(1 + {x}^{2}\right) + 2 x {\tan}^{-} 1 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x + {x}^{2} / \left(1 + {x}^{2}\right) + 2 x {\tan}^{-} 1 x$