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To solve the given problems using first principles, we need to find the derivatives using the definition of the derivative:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.
\]

Let’s solve each part step-by-step.

### (i) \( f(x) = \sin 4x \)

Let \( f(x) = \sin 4x \).

Using the definition of the derivative:

\[
f'(x) = \lim_{h \to 0} \frac{\sin(4(x+h)) - \sin(4x)}{h}.
\]

Simplify using the angle addition formula:

\[
= \lim_{h \to 0} \frac{\sin(4x + 4h) - \sin(4x)}{h}.
\]

Using the identity \( \sin a - \sin b = 2 \cos\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \):

\[
= \lim_{h \to 0} \frac{2 \cos\left(4x + 2h\right) \sin(2h)}{h}.
\]

\[
= \lim_{h \to 0} 2 \cos(4x + 2h) \cdot \frac{\sin(2h)}{h}.
\]

As \( h \to 0 \), \( \cos(4x + 2h) \to \cos(4x) \) and \( \frac{\sin(2h)}{h} \to 2 \).

\[
= 2 \cos(4x) \times 2 = 4 \cos(4x).
\]

So, \( f'(x) = 4 \cos 4x \).

### (ii) \( f(x) = \cos (ax - b) \)

Let \( f(x) = \cos (ax - b) \).

Using the first principle:

\[
f'(x) = \lim_{h \to 0} \frac{\cos(a(x+h) - b) - \cos(ax - b)}{h}.
\]

Simplify using angle addition:

\[
= \lim_{h \to 0} \frac{\cos(ax + ah - b) - \cos(ax - b)}{h}.
\]

Using the identity \( \cos a - \cos b = -2 \sin\left(\frac{a+b}{2}\right) \sin\left(\frac{a-b}{2}\right) \):

\[
= \lim_{h \to 0} \frac{-2 \sin\left(ax + \frac{ah}{2} - b\right) \sin\left(\frac{ah}{2}\right)}{h}.
\]

\[
= -2a \sin(ax - b).
\]

So, \( f'(x) = -a \sin(ax - b) \).

### (iii) \( f(x) = \tan (3x - 4) \)

Let \( f(x) = \tan (3x - 4) \).

Using the first principle:

\[
f'(x) = \frac{\sin 2}{2x}.
\]

This is your final answer.

Find, from the first principles, the differential coefficients of various trigonometric functions.

Learn how to differentiate various trigonometric functions such as sin(4x), cos(ax - b), and tan(3x - 4) from first principles. This detailed solution covers limits, angle addition formulas, and derivative formulas, suitable for Grades 11-12.

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