# How do you differentiate y= arctan(x - sqrt(1+x^2))?

Jun 20, 2016

d/dx tan^-1(x-sqrt(1+x^2)) = (1-(x/sqrt(1+x^2)))/(1+(x-sqrt(1+x^2))^2

#### Explanation:

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

Now, treat $x - \sqrt{1 + {x}^{2}}$ as $x$ in the above definition.

That would give us,
d/dx tan^-1(x-sqrt(1+x^2)) = 1/(1+(x-sqrt(1+x^2))^2

Don't forget the chain rule though!

The derivative of $x - \sqrt{1 + {x}^{2}}$ is $1 - \left(\frac{x}{\sqrt{1 + {x}^{2}}}\right)$

Multiplying the derivative would give us,

d/dx tan^-1(x-sqrt(1+x^2)) = (1-(x/sqrt(1+x^2)))/(1+(x-sqrt(1+x^2))^2