If f(x) = x², then find f' from the first principle. 2. Complete the following statement. Derivative of sum is of derivative and derivative of difference is of derivative. 3. If y = xn, then find dy/dx. 4. If e−2x+y = 3xy, then find dy/dx.

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

Substitute $f(x) = x^2$ and $f(x+h) = (x+h)^2$.

$$f' = \lim_{h \to 0} \frac{\boxed{(x+h)^2} - x^2}{h}$$

$$f' = \lim_{h \to 0} \frac{\boxed{x^2 + 2xh + h^2 - x^2}}{h}$$ $$f' = \lim_{h \to 0} \frac{\boxed{2xh + h^2}}{h}$$

$$f' = \lim_{h \to 0} \frac{h(2x + h)}{h}$$ $$f' = \lim_{h \to 0} \boxed{2x+h}$$

$$f' = 2x + 0$$ $$f' = \boxed{2x}$$

2. Complete the following statement. Derivative of sum is ______ of derivative and derivative of difference is ______ of derivative.

The derivative of a sum of functions is the sum of their derivatives. The derivative of a difference of functions is the difference of their derivatives.
Derivative of sum is sum of derivative and derivative of difference is difference of derivative.

3. If $y = x^n$, then find $\frac{dy}{dx}$.

Given $y = x^n$.

$$\frac{dy}{dx} = \boxed{nx^{n-1}}$$

4. If $e^{-2x+y} = 3xy$, then find $\frac{dy}{dx}$.

Using the chain rule for the left side and the product rule for the right side:

$$\frac{d}{dx}(e^{-2x+y}) = e^{-2x+y} \cdot \frac{d}{dx}(-2x+y) = e^{-2x+y}(-2 + \frac{dy}{dx})$$ $$\frac{d}{dx}(3xy) = \frac{d}{dx}(3x) \cdot y + 3x \cdot \frac{d}{dx}(y) = 3y + 3x\frac{dy}{dx}$$

Equating the derivatives, the given line in the problem is:

$$\frac{d}{dx}(-2x+y)e^{-2x+y} = 3xy' + \boxed{3y}$$

$$-2e^{-2x+y} + e^{-2x+y}\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$$ $$e^{-2x+y}\frac{dy}{dx} - 3x\frac{dy}{dx} = 3y + 2e^{-2x+y}$$

$$\frac{dy}{dx}(e^{-2x+y} - 3x) = 3y + 2e^{-2x+y}$$ $$\frac{dy}{dx} = \boxed{\frac{3y + 2e^{-2x+y}}{e^{-2x+y} - 3x}}$$