If \( y=\frac{\sin x}{1+\cos x} \), then \( \frac{d y}{d x}= \) \( \begin{array}{llll}\text { (a) } y \sin x & \text { (b) } y \cos x & \text { (c) } y \csc x & \text { (d) } y \sec x\end{array} \) If \( y=\tan x \), then \( \frac{d y}{d x}= \)

To find \(\frac{d y}{d x}\) for \(y = \frac{\sin x}{1 + \cos x}\), we can use the quotient rule. The derivative can be calculated as follows: \[ \frac{d y}{d x} = \frac{(1 + \cos x)(\cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x(1 + \cos x) + \sin^2 x}{(1 + \cos x)^2} \] This simplifies to \(\frac{1 + \cos x}{(1 + \cos x)^2} = y \sec^2 x\), leading us to: \(\frac{d y}{d x} = y \cos x\) Similarly, for \(y = \tan x\), the derivative \(\frac{d y}{d x} = \sec^2 x\) stems from the fact that the derivative of tangent is the square of secant. So: \( \frac{dy}{dx} = \sec^2 x\) For the fun factor: Did you know that mathematicians often have their own secret handshakes? It’s just a lot of high fives and the symbolic math wave! Also, the beauty of derivatives is that they represent rates of change. So next time you raise your hand in class to ask a question, think of it as finding the rate of change of your curiosity – will it skyrocket or plateau?