# How do you implicitly differentiate 18=(y-x)^2+y/x?

##### 1 Answer
May 8, 2016

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

#### Explanation:

An implicit function is a function, in which a variable is not explicitly derived from another variable, rather is defined implicitly by an implicit equation i.e. by associating one of the variables with the other.

In such cases, we differentiate, say $y$ normally, but as we derive w.r.t. say $x$, considering $y$ as a function of $x$, using chain rule, we multiply by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

In the given case $18 = {\left(y - x\right)}^{2} + \frac{y}{x}$, differentiating w.r,t, $x$, we get

$0 = 2 \left(y - x\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}} - 1\right) + \frac{x \frac{\mathrm{dy}}{\mathrm{dx}} - y}{x} ^ 2$ or

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

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

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

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

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

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