Q.Find the derivative of the following functions:

Concept: Product rule, quotient rule, and standard trigonometric derivatives.

We'll use $\frac{d}{dx}(\sin x) = \cos x$, $\frac{d}{dx}(\cos x) = -\sin x$, and derive the rest as needed.

(i) $\sin x \cos x$: Apply the product rule.

(ii) $\sec x = \frac{1}{\cos x}$: Use the quotient rule.

(iii) $5\sec x + 4\cos x$: Differentiate term by term.

(iv) $\operatorname{cosec} x = \frac{1}{\sin x}$: Use the quotient rule.

(v) $3\cot x + 5\operatorname{cosec} x$: Since $\cot x = \frac{\cos x}{\sin x}$, we have $\frac{d}{dx}(\cot x) = -\operatorname{cosec}^2 x$.

$\frac{d}{dx}(3\cot x + 5\operatorname{cosec} x) = -3\operatorname{cosec}^2 x - 5\operatorname{cosec} x \cot x$ …

The heart of this problem is knowing your trigonometric derivatives. While you can memorize formulas like $\frac{d}{dx}(\sec x) = \sec x \tan x$, understanding why they work—by writing secant and cosecant in terms of sine and cosine and applying the quotient rule—makes them stick. Once you have the building blocks, the rest is linearity: the derivative of a sum is the sum of derivatives, and constants pull out.

Let me work through each function systematically.

(i) $\sin x \cos x$

This is a product of two functions, so the product rule applies: $(uv)' = u'v + uv'$.

1. Let $u = \sin x$ and $v = \cos x$, so $u' = \cos x$ and $v' = -\sin x$.

2. Applying the product rule:

3. This can also be written using the double-angle identity as $\cos 2x$, though the expanded form is equally valid.

Derivative: $\cos^2 x - \sin^2 x$ (or $\cos 2x$)

(ii) $\sec x$

Secant is $\frac{1}{\cos x}$, so we use the quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$.

1. Let $u = 1$ and $v = \cos x$, so $u' = 0$ and $v' = -\sin x$.

2. Applying the quotient rule:

3. Rewrite in terms of secant and tangent:

Derivative: $\sec x \tan x$

(iii) $5\sec x + 4\cos x$

Use linearity and the derivatives we know.

1. The derivative of $5\sec x$ is $5 \cdot \sec x \tan x = 5\sec x \tan x$.

2. The derivative of $4\cos x$ is $4 \cdot (-\sin x) = -4\sin x$.

3. Adding them:

Derivative: $5\sec x \tan x - 4\sin x$

(iv) $\csc x$

Cosecant is $\frac{1}{\sin x}$, so again the quotient rule.

1. Let $u = 1$ and $v = \sin x$, so $u' = 0$ and $v' = \cos x$.

2. Applying the quotient rule:

3. Rewrite:

Derivative: $-\csc x \cot x$

(v) $3\cot x + 5\csc x$

We need the derivative of cotangent. Since $\cot x = \frac{\cos x}{\sin x}$, use the quotient rule.

1. The derivative of $3\cot x$ is $3 \cdot (-\csc^2 x) = -3\csc^2 x$. …