Derivatives are a fundamental concept in calculus, providing a mathematical framework for understanding rates of change. They are essential in various fields, from physics to economics, and form the basis for more advanced studies in mathematics.

**Understanding the Derivative**

Differentiation is a key concept in calculus, concerned with finding the rate at which a function changes at a point. It is about understanding the derivative, which is the slope of a curve at any given point.

**The Concept of a Derivative**

The derivative, denoted as $\frac{\mathrm{dy}}{\mathrm{dx}}$ or $f'(x)$, is the limit of the gradients of chords as they approach a specific point on a curve. To visualize this, consider the function $y = x^3$. As $x$ moves from $2$ to $2 + h$, the gradient of the chord provides an approximation of the derivative at $x = 2$.

**First and Second Derivatives**

**First Derivative**$( f'(x) )$

For $y = x^n$, the first derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = n x^{n-1}$. This derivative represents the rate of change of $y$ with respect to $x$.

**Second Derivative**$( f''(x) )$

This is the derivative of the derivative, $\frac{d^2 y}{d x^2}$. It provides information about the curvature or concavity of the graph.

**Dealing with Negative Powers and Constants**

**Negative Powers**

Transform expressions like $\frac{1}{x}$ into $x^{-1}$ and $\frac{1}{x^n}$ into $x^{-n}$ when $x$ is in the denominator or has a negative power.

**Constants**

The differentiation of constants is always zero. For instance, if $f(x) = 5$, then $f'(x) = 0$.

**Examples**

**Example 1: **

Find the derivative of $y = x^3$ at $x = 2$.

**Solution:**

1. Define the derivative function for $y = x^3$: $f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}.$

2. Plug in $x = 2$ into the derivative function: $f'(2) = \lim_{h \to 0} \frac{(2+h)^3 - 2^3}{h}$.

3. Expand and simplify: $(2+h)^3 = 8 + 12h + 6h^2 + h^3$, so, $f'(2) = \lim_{h \to 0} \frac{8 + 12h + 6h^2 + h^3 - 8}{h}$.

4. Cancel and compute the limit: Simplify to $f'(2) = \lim_{h \to 0} \frac{12h + 6h^2 + h^3}{h},$ cancel $h$ to get $f'(2) = \lim_{h \to 0} (12 + 6h + h^2)$, as $h$ approaches zero, $f'(2) = 12$.

Therefore, the derivative of $y = x^3$ at $x = 2$ is 12, indicating the slope of the tangent to the curve at this point is 12.

**Example 2: **

Find the second derivative of $y = x^3$.

**Solution:**

1. Calculate the first derivative using the power rule $( f'(x) = nx^{n-1} )): ( f'(x) = 3x^{3-1} = 3x^2.$

2. Derive the second derivative by differentiating $f'(x)$ again: $f''(x) = \frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x.$

Therefore, the second derivative of $y = x^3$ is $6x$, indicating how the slope of the tangent line changes as $x$ varies. As $x$ increases, the curve's steepness increases at a constant rate, which is essential for understanding the curve's concavity and inflection points.

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.