# What is the derivative of Tan^-1 (y/x)?

##### 3 Answers
Jul 6, 2018

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

#### Explanation:

$y = {\tan}^{-} 1 \left(\frac{y}{x}\right)$
$\tan y = \frac{y}{x}$
Using the chain rule on the left side:
$\frac{d}{\mathrm{dx}} \left(\tan y\right) = \left({\sec}^{2} y\right) y '$
Using the product rule on the right side:
$\frac{d}{\mathrm{dx}} \left(\frac{y}{x}\right) = - \frac{y}{x} ^ 2 + \frac{y '}{x}$
$\therefore \left({\sec}^{2} y\right) y ' = - \frac{y}{x} ^ 2 + \frac{y '}{x}$
$\left({\sec}^{2} y\right) y ' - \frac{y '}{x} = - \frac{y}{x} ^ 2$
$y ' \left({\sec}^{2} y - \frac{1}{x}\right) = - \frac{y}{x} ^ 2$
$y ' = - \frac{y}{{x}^{2} {\sec}^{2} y - x} = \frac{y}{x - {x}^{2} {\sec}^{2} y}$
$\therefore \frac{d}{\mathrm{dx}} \left({\tan}^{-} 1 \left(\frac{y}{x}\right)\right) = \frac{y}{x - {x}^{2} {\sec}^{2} y}$

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

#### Explanation:

$u = {\tan}^{-} 1 \left(\frac{y}{x}\right)$

This problem needs a slight prerequisite of chain rule, and quotient rule.
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{1 + {\left(\frac{y}{x}\right)}^{2}} \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{x} ^ 2$

$= {x}^{2} \frac{y - x \frac{\mathrm{dy}}{\mathrm{dx}}}{{x}^{2} + {y}^{2}}$

Thus,

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

Jul 6, 2018

It really depends upon what you are doing, and which independent variables matter to you.

Let $z \left(x , y\right) = {\tan}^{-} 1 \left(\frac{y}{x}\right) q \quad q \quad \implies \tan z = \frac{y}{x} q \quad \triangle$

Using the Quotient Rule and Implicit Differentiation on $\triangle$:

${\sec}^{2} z \setminus \mathrm{dz} = \frac{x \setminus \mathrm{dy} - y \setminus \mathrm{dx}}{x} ^ 2$

$\therefore \mathrm{dz} = \frac{{x}^{2}}{{x}^{2} + {y}^{2}} \cdot \setminus \frac{x \setminus \mathrm{dy} - y \setminus \mathrm{dx}}{x} ^ 2 q \quad \square$

$\implies \mathrm{dz} = \frac{x \setminus \mathrm{dy} - y \setminus \mathrm{dx}}{{x}^{2} + {y}^{2}}$

The partial derivatives are therefore:

• $\left\{\begin{matrix}{z}_{x} = - \frac{y}{{x}^{2} + {y}^{2}} \\ {z}_{y} = \frac{x}{{x}^{2} + {y}^{2}}\end{matrix}\right.$

But if $x$ is the independent variable, ie $y = y \left(x\right)$, then you have from $\square$:

• $\frac{\mathrm{dz}}{\mathrm{dx}} = \frac{x \setminus \frac{\mathrm{dy}}{\mathrm{dx}} - y \setminus \frac{\mathrm{dx}}{\mathrm{dx}}}{{x}^{2} + {y}^{2}}$

Which is the total derivative wrt $x$:

• $\frac{\mathrm{dz}}{\mathrm{dx}} = {z}_{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {z}_{x}$

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