OUESTION 1 1.1) Determine $$ f^{\prime}(x) $$ from first principles if $$ f(x)=x^{2}+3 $$ 1.2) Determine $$ \frac{d y}{d x} $$ if: 1.2.1) 1.2.2) $$ y=3 x^{2}+\frac{5}{x} $$ $$ y \sqrt[3]{x}+\left(2 x^{3}\right)^{2} $$

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**1.1) Computation of $$ f'(x) $$ for $$ f(x)=x^2+3 $$ from first principles**

We start with the definition of the derivative:
$$
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}
$$

**Step 1.** Compute $$ f(x+h) $$:
$$
f(x+h) = (x+h)^2 + 3 = x^2 + 2xh + h^2 + 3
$$

**Step 2.** Compute the difference $$ f(x+h)-f(x) $$:
$$
f(x+h)-f(x) = \bigl(x^2 + 2xh + h^2 + 3\bigr) - \bigl(x^2+3\bigr)= 2xh + h^2
$$

**Step 3.** Divide by $$ h $$:
$$
\frac{f(x+h)-f(x)}{h} = \frac{2xh + h^2}{h} = 2x + h
$$

**Step 4.** Take the limit as $$ h\to 0 $$:
$$
f'(x)=\lim_{h\to 0}(2x+h)=2x
$$

Thus, 
$$
f'(x)=2x.
$$

---

**1.2) Computation of $$ \frac{dy}{dx} $$**

**1.2.1) For $$ y=3x^2+\frac{5}{x} $$**

First, express $$ \frac{5}{x} $$ as $$ 5x^{-1} $$:
$$
y=3x^2+5x^{-1}.
$$

Differentiate term by term using the power rule:

- The derivative of $$ 3x^2 $$ is:
  $$
  \frac{d}{dx}\left(3x^2\right)= 3\cdot 2x^{2-1}=6x.
  $$

- The derivative of $$ 5x^{-1} $$ is:
  $$
  \frac{d}{dx}\left(5x^{-1}\right)= 5\cdot(-1)x^{-2}=-5x^{-2}.
  $$

Thus, 
$$
\frac{dy}{dx}=6x-5x^{-2} = 6x-\frac{5}{x^2}.
$$

---

**1.2.2) For $$ y=\sqrt[3]{x}+(2x^3)^2 $$**

First, rewrite each term with exponents:

- The cube root $$ \sqrt[3]{x} $$ is:
  $$
  x^{\frac{1}{3}}.
  $$

- Simplify $$(2x^3)^2$$:
  $$
  (2x^3)^2 = 2^2 \cdot (x^3)^2= 4x^6.
  $$

So the function becomes:
$$
y=x^{\frac{1}{3}}+4x^6.
$$

Differentiate term by term:

- The derivative of $$ x^{\frac{1}{3}} $$ is:
  $$
  \frac{d}{dx}\left(x^{\frac{1}{3}}\right)=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{-\frac{2}{3}}.
  $$

- The derivative of $$ 4x^6 $$ is:
  $$
  \frac{d}{dx}\left(4x^6\right)= 4\cdot 6x^{6-1}=24x^5.
  $$

Thus, 
$$
\frac{dy}{dx}=\frac{1}{3}x^{-\frac{2}{3}}+24x^5.
$$
**1.1) Derivative of $$ f(x) = x^2 + 3 $$ from first principles:** $$ f'(x) = 2x $$ **1.2) Derivatives:** 1.2.1) For $$ y = 3x^2 + \frac{5}{x} $$: $$ \frac{dy}{dx} = 6x - \frac{5}{x^2} $$ 1.2.2) For $$ y = \sqrt[3]{x} + (2x^3)^2 $$: $$ \frac{dy}{dx} = \frac{1}{3x^{\frac{2}{3}}} + 24x^5 $$

OUESTION 1 1.1) Determine \( f^{\prime}(x) \) from first principles if \( f(x)=x^{2}+3 \) 1.2) Determine \( \frac{d y}{d x} \) if: 1.2.1) 1.2.2) \( y=3 x^{2}+\frac{5}{x} \) \( y \sqrt[3]{x}+\left(2 x^{3}\right)^{2} \)

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