For the curve y=f(x) show that the gradient of the secant passing through the points (x, f(x)) and (x+h, f(x+h)) is (f(x+h)-f(x))/h. If y = 2x2 + 3, find the value of dy/dx and hence show that dy/dx = 4 at x=1. Differentiate y = 2x2 − x + 3 by first finding the value of dy/dx.

The gradient of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.
Given the points $(x, f(x))$ and $(x+h, f(x+h))$, we can substitute these into the formula:

$$m = \frac{f(x+h) - f(x)}{(x+h) - x}$$ $$m = \frac{f(x+h) - f(x)}{h}$$

This shows the required expression.
Answer for 4: $\boxed{\frac{f(x+h)-f(x)}{h}}$

Given $y = 2x^2 + 3$.
To find $\frac{dy}{dx}$, we differentiate each term with respect to $x$:

$$\frac{dy}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(3)$$

Using the power rule $\frac{d}{dx}(ax^n) = anx^{n-1}$ and $\frac{d}{dx}(c) = 0$:

$$\frac{dy}{dx} = 2(2x^{2-1}) + 0$$ $$\frac{dy}{dx} = 4x$$

Substitute $x=1$ into the derivative $\frac{dy}{dx} = 4x$:

$$\frac{dy}{dx} \Big|_{x=1} = 4(1)$$ $$\frac{dy}{dx} \Big|_{x=1} = 4$$

Answer for 5: $\boxed{\frac{dy}{dx} = 4x, \text{ and at } x=1, \frac{dy}{dx} = 4}$

Given $y = f(x) = 2x^2 - x + 3$.
First, find $f(x+\Delta x)$:

$$f(x+\Delta x) = 2(x+\Delta x)^2 - (x+\Delta x) + 3$$ $$f(x+\Delta x) = 2(x^2 + 2x\Delta x + (\Delta x)^2) - x - \Delta x + 3$$ $$f(x+\Delta x) = 2x^2 + 4x\Delta x + 2(\Delta x)^2 - x - \Delta x + 3$$

Next, find $\Delta y = f(x+\Delta x) - f(x)$:

$$\Delta y = (2x^2 + 4x\Delta x + 2(\Delta x)^2 - x - \Delta x + 3) - (2x^2 - x + 3)$$ $$\Delta y = 2x^2 + 4x\Delta x + 2(\Delta x)^2 - x - \Delta x + 3 - 2x^2 + x - 3$$ $$\Delta y = 4x\Delta x + 2(\Delta x)^2 - \Delta x$$

Now, find $\frac{\Delta y}{\Delta x}$:

$$\frac{\Delta y}{\Delta x} = \frac{4x\Delta x + 2(\Delta x)^2 - \Delta x}{\Delta x}$$

Factor out $\Delta x$ from the numerator:

$$\frac{\Delta y}{\Delta x} = \frac{\Delta x (4x + 2\Delta x - 1)}{\Delta x}$$ $$\frac{\Delta y}{\Delta x} = 4x + 2\Delta x - 1$$

To find the derivative $\frac{dy}{dx}$, we take the limit of $\frac{\Delta y}{\Delta x}$ as $\Delta x \to 0$:

$$\frac{dy}{dx} = \lim_{\Delta x \to 0} (4x + 2\Delta x - 1)$$

Substitute $\Delta x = 0$:

$$\frac{dy}{dx} = 4x + 2(0) - 1$$ $$\frac{dy}{dx} = 4x - 1$$

Answer for 6: $\boxed{\frac{\Delta y}{\Delta x} = 4x + 2\Delta x - 1, \text{ and } \frac{dy}{dx} = 4x - 1}$