# 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)?

Nov 22, 2015

$\frac{d}{\mathrm{dx}} \left[\tan \left(x y\right) - \cot \left({x}^{2}\right) = C\right]$

We will use implicit differentiation and the chain rule next.

$\stackrel{\text{product rule}}{\overbrace{\frac{d}{\mathrm{dx}} \left[x y\right]}} {\sec}^{2} \left(x y\right) - \frac{d}{\mathrm{dx}} \left[{x}^{2}\right] \cdot - {\csc}^{2} \left({x}^{2}\right) = 0$

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

We can now plug in $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$ and solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$, which will give us the slope of the tangent line at that point.

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

From here, it's just algebra.