# What is the derivative of arctan(x-1)?

May 19, 2016

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

#### Explanation:

We can differentiate this using implicit differentiation:

Let $y = \arctan \left(x - 1\right)$

$\implies \tan \left(y\right) = x - 1$

$\implies \frac{d}{\mathrm{dx}} \tan \left(y\right) = \frac{d}{\mathrm{dx}} \left(x - 1\right)$

$\implies {\sec}^{2} \left(y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 \left(y\right)$

Drawing a right triangle with an angle $y$ such that $\tan \left(y\right) = x - 1$, we can find that $\sec \left(y\right) = \sqrt{{x}^{2} - 2 x + 2}$. Plugging that in, we get our result:

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

Note that the same process shows that the formula for $\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{{x}^{2} + 1}$. If we already have that formula, then we can simply use the chain rule:

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

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