# How do you differentiate sin(arctan x)?

Sep 23, 2016

$\frac{1}{1 + {x}^{2}} \cos \left(\arctan x\right)$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(\sin \left(\arctan x\right)\right) = \cos \left(\arctan x\right) \frac{d}{\mathrm{dx}} \left(\arctan x\right) =$

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

Oct 5, 2016

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

#### Explanation:

Another method:

If we draw a right triangle with an angle $\theta$ such that $\tan \left(\theta\right) = x$, then $\theta = \arctan \left(x\right)$. Using that triangle as a reference, we will find that $\sin \left(\arctan \left(x\right)\right) = \sin \left(\theta\right) = \frac{x}{\sqrt{{x}^{2} + 1}}$.

We can now differentiate using the quotient rule
$\frac{d}{\mathrm{dx}} f \frac{x}{g} \left(x\right) = \frac{f ' \left(x\right) g \left(x\right) - f \left(x\right) g ' \left(x\right)}{g \left(x\right)} ^ 2$

$\frac{d}{\mathrm{dx}} \sin \left(\arctan \left(x\right)\right) = \frac{d}{\mathrm{dx}} \frac{x}{\sqrt{{x}^{2} + 1}}$

$= \frac{\sqrt{{x}^{2} + 1} - {x}^{2} / \sqrt{{x}^{2} + 1}}{{x}^{2} + 1}$

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

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

Note that this matches the other answer, as using the triangle, we can also see that $\cos \left(\arctan \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} + 1}}$