# What is the implicit derivative of 1=tanyx-y^2?

Sep 28, 2017

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

#### Explanation:

$\text{differentiate "color(blue)"implicitly with respect to x}$

$\text{differentiate "tan(yx)" using "color(blue)"chain/product rules}$

$0 = {\sec}^{2} \left(y x\right) \times \frac{d}{\mathrm{dx}} \left(y x\right) - 2 y \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$\textcolor{w h i t e}{\Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}} = y {\sec}^{2} \left(y x\right) + x {\sec}^{2} \left(y x\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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