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) \( y=3 x^{2}+\frac{5}{x} \) \( y \sqrt[3]{x}+\left(2 x^{3}\right)^{2} \)

**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, \[ \frac{dy}{dx}=\frac{1}{3}x^{-\frac{2}{3}}+24x^5. \]