Implicit differentiation at a point with trigonometry and a fraction.

Question

Find $\frac{dy}{dx}$ for $xcosy-2sin(\frac{y}{2})=0$ at (2,$\frac{\pi}{3}$)

I tried using the power rule and chain rule but I do not seem to solve the problem. Can someone tell me how to solve it?.

Answer 1

Let $y=y(x)$. Then $$-x \sin{(y)} \left( \frac{d}{d x} y\right) -\cos{\left( \frac{y}{2}\right) } \left( \frac{d}{d x} y\right) +\cos{(y)}=0$$ $$\frac{d}{d x} y=\frac{\cos{(y)}}{x \sin{(y)}+\cos{\left( \frac{y}{2}\right) }}$$ At $x=2,\;y=\pi/3$ we get $$\frac{dy}{dx}=\frac{1}{3\sqrt3}$$

Answer 2

$$x\,\cos(y) - 2\sin(\frac{y}{2}) = 0 ~~~~~~~\text{original problem}\\ d(x\,\cos(y) - 2\sin(\frac{y}{2})) = d(0) ~~~~~~~\text{differentiate both sides}\\ d(x\,\cos(y)) - d(2\sin(\frac{y}{2}))) = 0 ~~~~~~~\text{addition rule on the left, differential of 0 is 0 on right} \\ x\,d(\cos(y)) + \cos(y)\,d(x) - 2\,d(\sin(\frac{y}{2})) = 0 ~~~~~~~\text{product rule, constant multiplier rule} \\ -x\,\sin(y)\,dy + \cos(y)\,dx - 2\,\cos(\frac{y}{2})\,d(\frac{y}{2}) = 0 \\ \text{(cosine rule, basic differential of a variable, sine rule w/ chain rule)} \\ -x\,\sin(y)\,dy + \cos(y)\,dx - 2\,\cos(\frac{y}{2})\frac{dy}{2} = 0\\ \text{constant multiplier rule + basic differential of a variable}\\ -x\,\sin(y)\,dy + \cos(y)\,dx - \cos(\frac{y}{2})\,dy = 0 ~~~\text{(simplified)} $$ Now solve for $\frac{dy}{dx}$: $$ -x\,\sin(y)\,dy - \cos(\frac{y}{2})\,dy = - \cos(y)\,dx \\ \frac{dy}{dx} = \frac{\cos(y)}{x\,\sin(y) + \cos(\frac{y}{2})} $$ Note that this method uses differentials rather than implicit derivatives, but I find it more straightforward.