3. What is the geometrical meaning of differentiation? 4. Differentiate the following functions using the first principle. (a) $$ y=x^{2}+3 x-4 $$ (b) $$ y=\frac{3 x-4}{2 x+1} $$ (c) $$ y=\sqrt{x+3} $$ (d) $$ y=\frac{1}{\sqrt{3}} $$

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$$3.$$ Differentiation, geometrically, represents the slope of the tangent to a curve at a given point. In other words, it is the instantaneous rate of change of the function with respect to the variable, which can be visualized as the steepness or inclination of the curve at that point.

$$4.$$ We differentiate each function using the first principle. The first principle (definition of derivative) is given by

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

**(a) $$ y=x^{2}+3x-4 $$**

Let $$ f(x)=x^2+3x-4 $$. Then

$$
f(x+h)= (x+h)^2+3(x+h)-4.
$$

Expanding,

$$
\begin{aligned}
f(x+h) &= x^2+2xh+h^2+3x+3h-4,\[1mm]
f(x+h)-f(x) &= \bigl[x^2+2xh+h^2+3x+3h-4\bigr] - \bigl[x^2+3x-4\bigr] \
&= 2xh+h^2+3h.
\end{aligned}
$$

Thus,

$$
\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2+3h}{h} = 2x + h + 3.
$$

Taking the limit as $$ h \to 0 $$,

$$
f'(x)= 2x+3.
$$

---

**(b) $$ y=\frac{3x-4}{2x+1} $$**

Let $$ f(x)=\frac{3x-4}{2x+1} $$. Then

$$
f(x+h)= \frac{3(x+h)-4}{2(x+h)+1} = \frac{3x+3h-4}{2x+2h+1}.
$$

The difference is

$$
f(x+h)-f(x)= \frac{3x+3h-4}{2x+2h+1}-\frac{3x-4}{2x+1}.
$$

Combine the fractions:

$$
f(x+h)-f(x)= \frac{(3x+3h-4)(2x+1)-(3x-4)(2x+2h+1)}{(2x+2h+1)(2x+1)}.
$$

We expand both products:

1. Expand $$(3x+3h-4)(2x+1)$$:

$$
\begin{aligned}
(3x+3h-4)(2x+1) &= 3x(2x+1)+3h(2x+1)-4(2x+1)\[1mm]
&= 6x^2+3x+6xh+3h-8x-4\[1mm]
&= 6x^2+6xh -5x+3h-4.
\end{aligned}
$$

2. Expand $$(3x-4)(2x+2h+1)$$:

$$
\begin{aligned}
(3x-4)(2x+2h+1) &= 3x(2x+2h+1)-4(2x+2h+1)\[1mm]
&= 6x^2+6xh+3x-8x-8h-4\[1mm]
&= 6x^2+6xh-5x-8h-4.
\end{aligned}
$$

Subtracting the second expansion from the first:

$$
\begin{aligned}
\text{Numerator} &= \Bigl[6x^2+6xh-5x+3h-4\Bigr]-\Bigl[6x^2+6xh-5x-8h-4\Bigr]\[1mm]
&= (3h+8h)\[1mm]
&= 11h.
\end{aligned}
$$

Thus,

$$
\frac{f(x+h)-f(x)}{h}=\frac{11h}{h(2x+2h+1)(2x+1)} = \frac{11}{(2x+2h+1)(2x+1)}.
$$

Taking the limit as $$ h \to 0 $$, note that $$ 2x+2h+1 \to 2x+1 $$. Hence,

$$
f'(x)= \frac{11}{(2x+1)^2}.
$$

---

**(c) $$ y=\sqrt{x+3} $$**

Let $$ f(x)=\sqrt{x+3} $$. Then

$$
f(x+h)=\sqrt{x+h+3}.
$$

We have

$$
f(x+h)-f(x)= \sqrt{x+h+3} - \sqrt{x+3}.
$$

To simplify, multiply the numerator and denominator by the conjugate:

$$
\frac{\sqrt{x+h+3} - \sqrt{x+3}}{h} \times \frac{\sqrt{x+h+3} + \sqrt{x+3}}{\sqrt{x+h+3} + \sqrt{x+3}} = \frac{(x+h+3)-(x+3)}{h\left(\sqrt{x+h+3}+\sqrt{x+3}\right)}.
$$

Simplifying the numerator:

$$
(x+h+3) - (x+3)= h.
$$

So,

$$
\frac{h}{h\left(\sqrt{x+h+3}+\sqrt{x+3}\right)}=\frac{1}{\sqrt{x+h+3}+\sqrt{x+3}}.
$$

Taking the limit as $$ h \to 0 $$,

$$
f'(x) = \frac{1}{2\sqrt{x+3}}.
$$

---

**(d) $$ y=\frac{1}{\sqrt{3}} $$**

Since this is a constant function, we have

$$
f(x)=\frac{1}{\sqrt{3}} \quad \text{and} \quad f(x+h)=\frac{1}{\sqrt{3}}.
$$

Thus,

$$
f(x+h)-f(x)= 0.
$$

Then,

$$
f'(x)= \lim_{h\to 0}\frac{0}{h}= 0.
$$
Differentiation represents the slope of the tangent to a curve at a point, indicating the instantaneous rate of change of the function. **(a) $$ y=x^{2}+3x-4 $$:** $$ y' = 2x + 3 $$ **(b) $$ y=\frac{3x-4}{2x+1} $$:** $$ y' = \frac{11}{(2x+1)^2} $$ **(c) $$ y=\sqrt{x+3} $$:** $$ y' = \frac{1}{2\sqrt{x+3}} $$ **(d) $$ y=\frac{1}{\sqrt{3}} $$:** $$ y' = 0 $$

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