lf y=sqrt ((1+sinx)/(1-sinx)) show (dy)/(dx)=(1)/(1-sinx) y-e^tan(4+sin3x)+(1)/(cosx)+6iiy=(cosx)^x+2x^2

To find the derivative of $y$ with respect to $x$, we can use the chain rule and the quotient rule.<br /><br />First, let's rewrite $y$ as $y=\sqrt{\frac{1+\sin x}{1-\sin x}}$.<br /><br />Now, let's differentiate $y$ with respect to $x$:<br /><br />$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1+\sin x}{1-\sin x}}} \cdot \frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)$<br /><br />To find $\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right)$, we can use the quotient rule:<br /><br />$\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) = \frac{(1-\sin x) \cdot \cos x - (1+\sin x) \cdot (-\cos x)}{(1-\sin x)^2}$<br /><br />Simplifying this expression, we get:<br /><br />$\frac{d}{dx}\left(\frac{1+\sin x}{1-\sin x}\right) = \frac{\cos x + \sin x \cos x}{(1-\sin x)^2} = \frac{\cos x (1+\sin x)}{(1-\sin x)^2}$<br /><br />Now, substituting this back into our expression for $\frac{dy}{dx}$, we get:<br /><br />$\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1+\sin x}{1-\sin x}}} \cdot \frac{\cos x (1+\sin x)}{(1-\sin x)^2}$<br /><br />Simplifying this expression, we get:<br /><br />$\frac{dy}{dx} = \frac{\cos x (1+\sin x)}{2(1-\sin x)^{3/2}}$<br /><br />Therefore, $\frac{dy}{dx} \neq \frac{1}{1-\sin x}$.<br /><br />Regarding the second equation, it seems to be a complex equation involving $y$, $e^{\tan(4+\sin 3x)}$, $\frac{1}{\cos x}$, and $6i y$. It is not clear what the question is asking for, so it is not possible to provide a solution.