# How do you implicitly differentiate 4= xytan(x^2y) ?

$y ' = \frac{- y \cdot \tan \left({x}^{2} y\right) - 2 {x}^{2} {y}^{2} \cdot {\sec}^{2} \left({x}^{2} y\right)}{x \cdot \tan \left({x}^{2} y\right) + {x}^{3} y \cdot {\sec}^{2} \left({x}^{2} y\right)}$

#### Explanation:

from the given

$4 = x y \tan \left({x}^{2} y\right)$

differentiate both sides of the equation

$\frac{d}{\mathrm{dx}} \left(4\right) = \frac{d}{\mathrm{dx}} \left(x\right) \cdot y \cdot \tan \left({x}^{2} y\right) + \frac{d}{\mathrm{dx}} \left(y\right) \cdot x \cdot \tan \left({x}^{2} y\right) +$

xy*d/dx(tan(x^2 y)

$0 = y \cdot \tan \left({x}^{2} y\right) + x \cdot \tan \left({x}^{2} y\right) y ' + 2 {x}^{2} {y}^{2} \cdot {\sec}^{2} \left({x}^{2} y\right) + {x}^{3} y$${\sec}^{2} \left({x}^{2} y\right) y '$

$- y \tan \left({x}^{2} y\right) - 2 {x}^{2} {y}^{2} {\sec}^{2} \left({x}^{2} y\right) = \left(x \tan \left({x}^{2} y\right) + {x}^{3} y {\sec}^{2} \left({x}^{2} y\right)\right) y '$

$y ' = \frac{- y \cdot \tan \left({x}^{2} y\right) - 2 {x}^{2} {y}^{2} \cdot {\sec}^{2} \left({x}^{2} y\right)}{x \cdot \tan \left({x}^{2} y\right) + {x}^{3} y \cdot {\sec}^{2} \left({x}^{2} y\right)}$