# How do you implicitly differentiate -1=-y^2x-2xy+e^x ?

##### 1 Answer
Jan 17, 2017

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

#### 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 $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Further, as the question also involves product of variables, such as $- {y}^{2} x$ and $- 2 x y$, we have to use Product Rule too. Product rule states if $f \left(x\right) = g \left(x\right) h \left(x\right)$

then $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{dg}}{\mathrm{dx}} \times h \left(x\right) + \frac{\mathrm{dh}}{\mathrm{dx}} \times g \left(x\right)$

Hence using the two

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

$= - {y}^{2} \times 1 - x \times 2 y \times \frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{2} - 2 x y \frac{\mathrm{dy}}{\mathrm{dx}}$ and

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

As $\frac{d}{\mathrm{dx}} {e}^{x} = {e}^{x}$

implicitly differentiating $- 1 = - {y}^{2} x - 2 x y + {e}^{x}$ we have

$0 = - {y}^{2} - 2 x y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x \frac{\mathrm{dy}}{\mathrm{dx}} - 2 y + {e}^{x}$ or

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

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