Q.Compute the derivative of

Concept: Derivative at a Point — we apply standard differentiation rules directly.

1. For $f(x) = \sin 2x$, use the chain rule: $\frac{d}{dx} \sin(2x) = \cos(2x) \cdot \frac{d}{dx}(2x) = \cos(2x) \cdot 2 = 2\cos 2x$.
2. For $g(x) = \cot x$, recall that $\cot x = \frac{\cos x}{\sin x}$. …

The Core Idea: Derivative at a Point vs. Derivative as a Function

When we say "compute the derivative," we mean find the derivative function — a formula that gives the slope of the tangent line at any point $x$. The two functions here, $\sin 2x$ and $\cot x$, are built from simpler pieces. The key is to see each as a composition or ratio of basic functions whose derivatives we already know.

For $\sin 2x$, think: "sine of double the angle." That's a chain: $2x$ inside, $\sin$ outside. For $\cot x$, think: "cosine over sine." That's a quotient: $\frac{\cos x}{\sin x}$.

Let's work through each.

(i) $f(x) = \sin 2x$

1. Recognize the structure.

$f(x) = \sin(2x)$ is a composite function. The outer function is $\sin(u)$, and the inner function is $u = 2x$. Whenever you have a function "wrapped around" another, the chain rule applies.

2. Differentiate the outer function, leaving the inside alone.

The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$. So:

3. Multiply by the derivative of the inner function.

The inner function is $g(x) = 2x$. Its derivative is simply $2$:

4. Apply the chain rule.

Multiply the two results:

5. Final result for (i):

(ii) $g(x) = \cot x$

1. Rewrite in terms of sine and cosine.

The definition of cotangent is:

This is a quotient of two functions: numerator $N(x) = \cos x$, denominator $D(x) = \sin x$.

2. Recall the quotient rule.

For $g(x) = \frac{N(x)}{D(x)}$, the derivative is:

$g'(x) = \frac{N'(x) D(x) - N(x) D'(x)}{[D(x)]^2}$ …