# Question #ea7ac

Dec 12, 2017

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

#### Explanation:

We will use implicit differentiation.
$y = {\tan}^{-} 1 \left(x\right)$

We can take $\tan$ on both sides (since it'll eliminate the inverse tan, and we already know how to differentiate it):
$\tan \left(y\right) = \cancel{\tan} \left(\cancel{{\tan}^{-} 1} \left(x\right)\right)$

$\tan \left(y\right) = x$

Now we can differentiate both sides with respect to $x$:
$\frac{d}{\mathrm{dx}} \left(\tan \left(y\right)\right) = \frac{\mathrm{dx}}{\mathrm{dx}}$

To work out the left hand side we need to be careful. Since we're differentiating with respect to $x$ on something involving $y$ (which is a function of $x$), the chain rule requires us to also multiply by the derivative of the inside:
${\sec}^{2} \left(y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Now we can divide both sides by ${\sec}^{2} \left(y\right)$ to isolate the derivative:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\sec} ^ 2 \left(y\right)$

Now, we want the derivative in terms of $x$, so we substitute in for $y = {\tan}^{-} 1 \left(x\right)$:
$\frac{1}{\sec} ^ 2 \left({\tan}^{-} 1 \left(x\right)\right)$

To simplify, we'll think of a right triangle with adjacent $1$, opposite $x$ (this is from the inverse tan, tangent is opposite over adjacent) and hypotenuse $\sqrt{{x}^{2} + 1}$ (pythagorean theorem).

Since secant is just hypotenuse over adjacent, we get:
$\frac{1}{\sec} ^ 2 \left({\tan}^{-} 1 \left(x\right)\right) = \frac{1}{\frac{\sqrt{{x}^{2} + 1}}{1}} ^ 2 = \frac{1}{{x}^{2} + 1}$

So by this, we've figured out that the derivative of ${\tan}^{-} 1 \left(x\right)$ is:
$\frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(x\right)\right) = \frac{1}{{x}^{2} + 1}$