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CIE A-Level Maths Study Notes

1.7.1 Understanding the Derivative

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 dydx\frac{\mathrm{dy}}{\mathrm{dx}} or f(x)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=x3y = x^3. As xx moves from 22 to 2+h2 + h, the gradient of the chord provides an approximation of the derivative at x=2x = 2.

First and Second Derivatives

  • First Derivative (f(x))( f'(x) )

For y=xny = x^n, the first derivative is dydx=nxn1\frac{\mathrm{dy}}{\mathrm{dx}} = n x^{n-1}. This derivative represents the rate of change of yy with respect to xx.

  • Second Derivative (f(x))( f''(x) )

This is the derivative of the derivative, d2ydx2\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 1x\frac{1}{x} into x1x^{-1} and 1xn\frac{1}{x^n} into xnx^{-n} when xx is in the denominator or has a negative power.

  • Constants

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

Examples

Example 1:

Find the derivative of y=x3y = x^3 at x=2x = 2.

Solution:

1. Define the derivative function for y=x3y = x^3: f(x)=limh0(x+h)3x3h.f'(x) = \lim_{h \to 0} \frac{(x+h)^3 - x^3}{h}.

2. Plug in x=2x = 2 into the derivative function: f(2)=limh0(2+h)323hf'(2) = \lim_{h \to 0} \frac{(2+h)^3 - 2^3}{h}.

3. Expand and simplify: (2+h)3=8+12h+6h2+h3(2+h)^3 = 8 + 12h + 6h^2 + h^3, so, f(2)=limh08+12h+6h2+h38hf'(2) = \lim_{h \to 0} \frac{8 + 12h + 6h^2 + h^3 - 8}{h}.

4. Cancel and compute the limit: Simplify to f(2)=limh012h+6h2+h3h, f'(2) = \lim_{h \to 0} \frac{12h + 6h^2 + h^3}{h}, cancel hh to get f(2)=limh0(12+6h+h2)f'(2) = \lim_{h \to 0} (12 + 6h + h^2), as hh approaches zero, f(2)=12f'(2) = 12.

Therefore, the derivative of y=x3y = x^3 at x=2x = 2 is 12, indicating the slope of the tangent to the curve at this point is 12.

derivatives

Example 2:

Find the second derivative of y=x3y = x^3.

Solution:

1. Calculate the first derivative using the power rule (f(x)=nxn1)):(f(x)=3x31=3x2.( f'(x) = nx^{n-1} )): ( f'(x) = 3x^{3-1} = 3x^2.

2. Derive the second derivative by differentiating f(x)f'(x) again: f(x)=ddx(3x2)=32x21=6x.f''(x) = \frac{d}{dx}(3x^2) = 3 \cdot 2x^{2-1} = 6x.

Therefore, the second derivative of y=x3y = x^3 is 6x6x, indicating how the slope of the tangent line changes as xx varies. As x x increases, the curve's steepness increases at a constant rate, which is essential for understanding the curve's concavity and inflection points.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

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.

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