What is the slope of the tangent line of xarctan(y/pi)= C , where C is an arbitrary constant, at (pi/3,pi/3)?

Mar 19, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{10}{9} \arctan \left(1 / 3\right)$

Explanation:

Taking the derivative, we first use the product rule:

$\left(\frac{d}{\mathrm{dx}} x\right) \arctan \left(\frac{y}{\pi}\right) + x \left(\frac{d}{\mathrm{dx}} \arctan \left(\frac{y}{\pi}\right)\right) = 0$

The typical arctangent derivative is $\frac{d}{\mathrm{dx}} \arctan \left(x\right) = \frac{1}{1 + {x}^{2}}$, so here we apply the chain rule:

$\arctan \left(\frac{y}{\pi}\right) + x \left(\frac{1}{1 + {\left(\frac{y}{\pi}\right)}^{2}}\right) \left(\frac{d}{\mathrm{dx}} \frac{y}{\pi}\right) = 0$

The derivative with respect to $x$ of $\frac{y}{\pi}$, which is the linear equation $\frac{1}{\pi} y$, so its derivative is $\frac{1}{\pi} \frac{\mathrm{dy}}{\mathrm{dx}}$:

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

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$:

$\arctan \left(1 / 3\right) + \frac{\pi / 3}{1 + \left({\pi}^{2} / 9\right) / {\pi}^{2}} \left(\frac{1}{\pi / 3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\arctan \left(1 / 3\right) + \left(\frac{1}{1 + 1 / 9}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\arctan \left(1 / 3\right) + \frac{9}{10} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{10}{9} \arctan \left(1 / 3\right)$