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Let's go through the problem step by step.

### a. Given \( y = x^2 - 1 \), find \( \frac{dy}{dx} \) from first principles.

**Solution:**

To differentiate \( y = x^2 - 1 \) using first principles, we use the definition of the derivative:

\[
\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

Substituting \( f(x) = x^2 - 1 \):

\[
f(x+h) = (x+h)^2 - 1 = x^2 + 2xh + h^2 - 1
\]

Now,

\[
f(x+h) - f(x) = (x^2 + 2xh + h^2 - 1) - (x^2 - 1) = 2xh + h^2
\]

Therefore,

\[
\frac{dy}{dx} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x + h) = 2x
\]

So, \( \frac{dy}{dx} = 2x \).

### b. Hence, the gradient of the tangent to \( y = x^2 - 1 \) at the point where \( x = 1 \).

To find the gradient at \( x = 1 \):

\[
\frac{dy}{dx} \text{ at } x = 1 = 2(1) = 2
\]

Thus, the gradient of the tangent at \( x = 1 \) is \( 2 \).

### c. For what value of \( x \) is the gradient of the tangent to \( y = x^2 - 1 \) equal to \( -2 \)?

We set \( \frac{dy}{dx} = -2 \):

\[
2x = -2 \quad \Rightarrow \quad x = -1
\]

So, the gradient of the tangent is \( -2 \) when \( x = -1 \).

### d. The graph of \( y = x^3 - 3x \) is shown on the diagram.

**i.** Find \( \frac{dy}{dx} \).

To differentiate \( y = x^3 - 3x \):

\[
\frac{dy}{dx} = 3x^2 - 3
\]

**ii.** Hence, show that the illustrated tangents are parallel.

The tangents are parallel if they have the same gradient. For the tangent at a point \( x = a \):

\[
\frac{dy}{dx} = 3a^2 - 3
\]

Given two points \( x = a \) and \( x = b \), if the tangents are parallel, then:

\[
3a^2 - 3 = 3b^2 - 3 \quad \Rightarrow \quad a^2 = b^2 \quad \Rightarrow \quad a = \pm b
\]

The tangents at \( x = a \) and \( x = -a \) are parallel.

### Questions for Further Exploration:
1. How would the derivative change if the function was \( y = 2x^2 - 3x + 4 \)?
2. What is the geometric interpretation of the derivative at a point?
3. How can we verify that two curves intersect perpendicularly?
4. What is the second derivative, and how does it relate to the concavity of the graph?
5. Can you find the equation of the tangent line to \( y = x^2 - 1 \) at \( x = 2 \)?

### Tip:
Always remember that finding the derivative from first principles involves simplifying the difference quotient before taking the limit. This ensures accuracy in your calculations.

Learn how to differentiate y = x^2 - 1 using first principles and find the gradient of the tangent at x = 1 and x = -1. This problem covers differentiation, tangent lines, and first principles in calculus, suitable for high school students.

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