# How do you implicitly differentiate  sin(y-x)^2-y=2?

##### 1 Answer
Apr 15, 2018

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

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

Hence differentiating $\sin {\left(y - x\right)}^{2} - y = 2$, we get

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

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

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

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} \left[2 \left(y - x\right) \cos {\left(y - x\right)}^{2} - 1\right] = 2 \left(y - x\right) \cos {\left(y - x\right)}^{2}$

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