Do question A ,B,C

Here are the solutions for questions A, B, and C.

A) If $f(x) = 3x^2 - 2x + 1$, find $f'(x)$ from first principles.

Step 1: Write down the definition of the derivative from first principles and find $f(x+h)$. The definition of the derivative is: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ Given $f(x) = 3x^2 - 2x + 1$, we find $f(x+h)$: $f(x+h) = 3(x+h)^2 - 2(x+h) + 1$ $f(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h + 1$ $f(x+h) = 3x^2 + 6xh + 3h^2 - 2x - 2h + 1$

Step 2: Calculate $f(x+h) - f(x)$. $f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h + 1) - (3x^2 - 2x + 1)$ $f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 2x - 2h + 1 - 3x^2 + 2x - 1$ $f(x+h) - f(x) = 6xh + 3h^2 - 2h$

Step 3: Divide by $h$ and simplify. $\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2 - 2h}{h}$ $\frac{f(x+h) - f(x)}{h} = \frac{h(6x + 3h - 2)}{h}$ $\frac{f(x+h) - f(x)}{h} = 6x + 3h - 2$

Step 4: Take the limit as $h \to 0$. $f'(x) = \lim_{h \to 0} (6x + 3h - 2)$ $f'(x) = 6x + 3(0) - 2$ $\boxed{f'(x) = 6x - 2}$

B) Differentiate $y = \frac{3x^2 - 2x + 1}{\sqrt{x}}$ with respect to $x$.

Step 1: Rewrite the function by expressing $\sqrt{x}$ as $x^{1/2}$ and dividing each term in the numerator. $y = \frac{3x^2}{x^{1/2}} - \frac{2x}{x^{1/2}} + \frac{1}{x^{1/2}}$ $y = 3x^{2 - 1/2} - 2x^{1 - 1/2} + x^{-1/2}$ $y = 3x^{3/2} - 2x^{1/2} + x^{-1/2}$

Step 2: Differentiate each term using the power rule, $\frac{d}{dx}(ax^n) = anx^{n-1}$. $\frac{dy}{dx} = \frac{d}{dx}(3x^{3/2}) - \frac{d}{dx}(2x^{1/2}) + \frac{d}{dx}(x^{-1/2})$ $\frac{dy}{dx} = 3 \left(\frac{3}{2}\right) x^{\frac{3}{2} - 1} - 2 \left(\frac{1}{2}\right) x^{\frac{1}{2} - 1} + \left(-\frac{1}{2}\right) x^{-\frac{1}{2} - 1}$ $\frac{dy}{dx} = \frac{9}{2} x^{1/2} - 1 x^{-1/2} - \frac{1}{2} x^{-3/2}$

Step 3: Simplify the expression. \boxed{\frac{dy{dx} = \frac{9}{2} \sqrt{x} - \frac{1}{\sqrt{x}} - \frac{1}{2x\sqrt{x}}}}

C) Find the equation of the tangent to the curve $y = x^3 - 2x^2 + 3x - 1$ at the point where $x=2$.

Step 1: Find the y-coordinate of the point of tangency. Substitute $x=2$ into the equation of the curve: $y = (2)^3 - 2(2)^2 + 3(2) - 1$ $y = 8 - 2(4) + 6 - 1$ $y = 8 - 8 + 6 - 1$ $y = 5$ The point of tangency is $(2, 5)$.

Step 2: Find the derivative $\frac{dy}{dx}$ to determine the slope function. $y = x^3 - 2x^2 + 3x - 1$ $\frac{dy}{dx} = 3x^2 - 2(2x) + 3(1) - 0$ $\frac{dy}{dx} = 3x^2 - 4x + 3$

Step 3: Evaluate the derivative at $x=2$ to find the slope ($m$) of the tangent. $m = 3(2)^2 - 4(2) + 3$ $m = 3(4) - 8 + 3$ $m = 12 - 8 + 3$ $m = 7$

Step 4: Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, to find the equation of the tangent. Using the point $(x_1, y_1) = (2, 5)$ and slope $m=7$: $y - 5 = 7(x - 2)$ $y - 5 = 7x - 14$ $y = 7x - 14 + 5$ $\boxed{y = 7x - 9}$