# How do you take the derivative of tan(x - y) = y/(1 + x^2) ?

Sep 16, 2015

Have a look, but check my maths.

#### Explanation:

In this case you'll need to derive implicitly considering that here $y$ represents a function of $x$ (a little bit...difficult to work out!) and we derive also $y$:
so if you want, say, to derive $y$ you get $1 \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

so we get (using the Chain Rule on the left and the Quotient Rule on the right):

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

rearrange to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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

Multiply the non-fractions by ${\cos}^{2} \frac{x - y}{\cos} ^ 2 \left(x - y\right)$ to get common denominators, cancel out those denominators, then divide by the resultant left-side bracketed contents:

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