The problem shown involves differentiating $y = \tan x$, and the hint suggests rewriting $\tan x$ using the identity $\tan x = \frac{\sin x}{\cos x}$.

Let's break this down step-by-step:

Step 1: Rewrite the function using the hint

$y = \frac{\sin x}{\cos x}$

Step 2: Differentiate using the quotient rule

The quotient rule states: $\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ Here, $f(x) = \sin x$ and $g(x) = \cos x$.

1. Derivative of $f(x) = \sin x$ is $f'(x) = \cos x$.
2. Derivative of $g(x) = \cos x$ is $g'(x) = -\sin x$.

Applying the quotient rule: $\frac{d}{dx} \left( \frac{\sin x}{\cos x} \right) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{(\cos x)^2}$ This simplifies to: $\frac{\cos^2 x + \sin^2 x}{\cos^2 x}$ Using the Pythagorean identity $\cos^2 x + \sin^2 x = 1$: $\frac{1}{\cos^2 x}$

Thus, the derivative of $y = \tan x$ is: $\frac{d}{dx} \tan x = \sec^2 x$

However, the answer provided at the top shows $2 \sin x \cos x$, which is another form related to trigonometric identities but does not directly seem related to the derivative of $\tan x$. The differentiation result for $y = \tan x$ should be $\sec^2 x$, not $2 \sin x \cos x$.

Would you like more details on the answer provided in the image? Or any additional clarification?

Here are 5 questions to further explore related concepts:

1. What are the steps for simplifying $2 \sin x \cos x$ into another form?
2. Can $\tan x = \frac{\sin x}{\cos x}$ be used in other calculus applications?
3. How does the quotient rule work for more complex trigonometric functions?
4. How would the result change if we were differentiating $\cot x$?
5. What are some practical applications of differentiating trigonometric functions?

Tip: When working with trigonometric identities, recognizing alternative forms can simplify integration and differentiation.