# What is the slope of the tangent line of tan(xy)/cot(x^2)= C , where C is an arbitrary constant, at (pi/3,pi/3)?

Feb 15, 2016

I get $m = - 3$.

#### Explanation:

To use product rule instead of quotient, rewrite

$\tan \left(x y\right) \tan \left({x}^{2}\right) = C$

Differentiate using the product and chain rules.

(I use $\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$ for the product rule.)

${\sec}^{2} \left(x y\right) \left[y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right] \tan \left({x}^{2}\right) + \tan \left(x y\right) {\sec}^{2} \left({x}^{2}\right) \left[2 x\right] = 0$

At $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$, we have

${\sec}^{2} \left({\pi}^{2} / 9\right) \left[\frac{\pi}{3} + \frac{\pi}{3} \frac{\mathrm{dy}}{\mathrm{dx}}\right] \tan \left({\pi}^{2} / 9\right) + \tan \left({\pi}^{2} / 9\right) {\sec}^{2} \left({\pi}^{2} / 9\right) \left[2 \frac{\pi}{3}\right] = 0$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \pi {\sec}^{2} \left({\pi}^{2} / 9\right) \tan \left({\pi}^{2} / 9\right)}{\frac{\pi}{3} {\sec}^{2} \left({\pi}^{2} / 9\right) \tan \left({\pi}^{2} / 9\right)} = - 3$