# What is the slope of the tangent line of sinx-sin(y^2)/x= C , where C is an arbitrary constant, at (pi/3,pi/2)?

##### 1 Answer
Nov 22, 2015

$\frac{{\pi}^{2} + 18 \sin \left({\pi}^{2} / 4\right)}{6 {\pi}^{2} \sin \left({\pi}^{2} / 4\right)}$

#### Explanation:

The constant $C$
If we know that $\left(\frac{\pi}{3} , \frac{\pi}{2}\right)$ is a point on the graph, then $C = \sin \left(\frac{\pi}{3}\right) - \sin \frac{{\left(\frac{\pi}{2}\right)}^{2}}{\frac{\pi}{3}} = \frac{\sqrt{3}}{2} - \frac{3 \sin \left({\left(\frac{\pi}{2}\right)}^{2}\right)}{\pi}$.
So $C$ is determined by the values of $x$ and $y$. It is not arbitrary. (It is not independent of the values of the variables.)

Slope of Tangent
Regardless of that, we can find the slope of the tangent line by implicit differentiation.

$\frac{d}{\mathrm{dx}} \left(\sin x - \sin \frac{{y}^{2}}{x}\right) = \frac{d}{\mathrm{dx}} \left(C\right)$

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

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

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

The slope at $\left(\frac{\pi}{3} , \frac{\pi}{2}\right)$, is therefore

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(\frac{\pi}{3}\right)}^{2} \cos \left(\frac{\pi}{3}\right) + \sin \left({\left(\frac{\pi}{2}\right)}^{2}\right)}{2 \left(\frac{\pi}{3}\right) \left(\frac{\pi}{2}\right) \sin \left({\left(\frac{\pi}{2}\right)}^{2}\right)}$

$= \frac{{\pi}^{2} + 18 \sin \left({\pi}^{2} / 4\right)}{6 {\pi}^{2} \sin \left({\pi}^{2} / 4\right)}$