# How do you find the derivative of tan(x/y)=x+y?

##### 2 Answers
Dec 24, 2017

$- \frac{{y}^{2} - y {\sec}^{2} \left(\frac{x}{y}\right)}{x {\sec}^{2} \left(\frac{x}{y}\right) + {y}^{2}}$

#### Explanation:

$\text{differentiate "tan(x/y)" using the "color(blue)"chain rule}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(\tan \left(\frac{x}{y}\right)\right)$

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

$\text{differentiate "x/y" using the "color(blue)"quotient rule}$

$= {\sec}^{2} \left(\frac{x}{y}\right) \times \frac{y - x . \frac{\mathrm{dy}}{\mathrm{dx}}}{y} ^ 2$

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

$\text{returning to the original}$

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

$\Rightarrow y {\sec}^{2} \left(\frac{x}{y}\right) - x {\sec}^{2} \left(\frac{x}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} + {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(- x {\sec}^{2} \left(\frac{x}{y}\right) - {y}^{2}\right) = {y}^{2} - y {\sec}^{2} \left(\frac{x}{y}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{y}^{2} - y {\sec}^{2} \left(\frac{x}{y}\right)}{x {\sec}^{2} \left(\frac{x}{y}\right) + {y}^{2}}$

Dec 24, 2017

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

#### Explanation:

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

You need the chain rule on the tangent part:

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

Distribute on the left side:

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

Move everything with a $\frac{\mathrm{dy}}{\mathrm{dx}}$ to the left and everything without to the right:

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

Factor out the $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

Get $\frac{\mathrm{dy}}{\mathrm{dx}}$ by itself by dividing both sides by that coefficient:

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

Get a common denominator in both the numerator and the denominator:

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

Simplify the complex fraction (I think of it as keep, change, turn):

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

I factored the negative out of the denominator and distributed it to the numerator:

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

I decided to factor the common $y$ out of the numerator:

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