# How do you implicitly derive e^cos y=x^3 arctan y?

##### 1 Answer
Mar 27, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} \arctan y}{\sin y {e}^{\cos} y + {x}^{3} / \left(1 + {y}^{2}\right)}$

#### Explanation:

Implicit differentiation is a special case of the chain rule for derivatives. Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$. However, some functions y are written implicitly as functions of $x$. So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

${e}^{\cos} y = {x}^{3} \arctan y$

then taking differential

${e}^{\cos} y \times \frac{d}{\mathrm{dy}} \cos y \times \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \arctan y + {x}^{3} \times \frac{1}{1 + {y}^{2}} \times \frac{\mathrm{dy}}{\mathrm{dx}}$

or $- \sin y {e}^{\cos} y \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} \arctan y + {x}^{3} / \left(1 + {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

or $\left(\sin y {e}^{\cos} y + {x}^{3} / \left(1 + {y}^{2}\right)\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2} \arctan y$

and $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} \arctan y}{\sin y {e}^{\cos} y + {x}^{3} / \left(1 + {y}^{2}\right)}$