To find the maximum or minimum value of the function $f(x) = \frac{\sin x}{1 + \tan x}$, we can analyze it using calculus by differentiating and finding the critical points. Here's a step-by-step approach:

Step 1: Express the function in terms of sine and cosine

Recall that: $\tan x = \frac{\sin x}{\cos x}$ Thus, the function becomes: $f(x) = \frac{\sin x}{1 + \frac{\sin x}{\cos x}} = \frac{\sin x}{\frac{\cos x + \sin x}{\cos x}} = \frac{\sin x \cdot \cos x}{\cos x + \sin x}$ Simplifying: $f(x) = \frac{\sin x \cdot \cos x}{\cos x + \sin x}$

Step 2: Differentiate the function

Now, differentiate $f(x)$ with respect to $x$ using the quotient rule.

$\text{Let } f(x) = \frac{g(x)}{h(x)} \text{, where } g(x) = \sin x \cdot \cos x \text{ and } h(x) = \cos x + \sin x$

The quotient rule states: $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$

Differentiate $g(x) = \sin x \cdot \cos x$:

$g'(x) = \cos^2 x - \sin^2 x = \cos(2x)$

Differentiate $h(x) = \cos x + \sin x$:

$h'(x) = -\sin x + \cos x$

Now, substitute these into the quotient rule formula: $f'(x) = \frac{\cos(2x)(\cos x + \sin x) - (\sin x \cdot \cos x)(-\sin x + \cos x)}{(\cos x + \sin x)^2}$

This derivative will help us find the critical points where $f'(x) = 0$, which correspond to the maximum or minimum values of the function.

Would you like me to proceed with simplifying the derivative or finding the critical points?

Here are 5 related questions:

1. What is the quotient rule, and how is it applied to find derivatives?
2. How do you find the critical points of a function using derivatives?
3. Can trigonometric identities simplify the differentiation process?
4. How can you identify if a critical point is a maximum or minimum?
5. What role do trigonometric transformations play in optimizing functions?

Tip: Always simplify a function before differentiating to make the process smoother.