First and Second Derivatives
Differentiation is the process of finding the instantaneous rate of change of a function. This is geometrically interpreted as finding the gradient of the tangent to the function's graph at any given point.
Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- The derivative, written as f'(x) or dy/dx, gives a formula for the gradient of the curve y = f(x).
- To find the gradient at a specific point, substitute the x-coordinate of that point into the derivative expression.
- The derivative represents the instantaneous rate of change. For example, if a function describes distance vs. time, its derivative gives the instantaneous velocity.
- The second derivative, f''(x) or d²y/dx², describes the rate of change of the gradient. It tells you how the slope of the curve is changing.
- A positive second derivative means the gradient is increasing (the curve is convex, or 'bending upwards'). A negative second derivative means the gradient is decreasing (the curve is concave, or 'bending downwards').
Formulae
f'(x) or dy/dx Standard notations for the first derivative of a function y = f(x) with respect to x.
f''(x) or d²y/dx² Standard notations for the second derivative of a function y = f(x) with respect to x.
Definitions
- Derivative (First Derivative)
- A function, denoted f'(x) or dy/dx, that gives the gradient of the tangent to a curve y = f(x) for any value of x.
- Tangent
- A straight line that touches a curve at a single point and has the same gradient as the curve at that point.
- Second Derivative
- The derivative of the first derivative, denoted f''(x) or d²y/dx². It measures the rate of change of the gradient.
- Rate of Change
- How one quantity changes in relation to another. The derivative gives the instantaneous rate of change, not the average rate of change over an interval.
Worked example
A curve is defined by the equation y = 2x³ + 3x² - 12x + 1. The tangent to this curve at a point P has a gradient of 0. Given that the x-coordinate of P is positive, what is the y-coordinate of P?
- 1
The gradient of the tangent is given by the first derivative, dy/dx.
- 2
First, find the derivative of the function:
dy/dx = 6x² + 6x - 12 - 3
We are told the gradient is 0, so set the derivative equal to 0:
6x² + 6x - 12 = 0.
- 4
Solve this quadratic equation for x.
Divide by 6 to simplify:
x² + x - 2 = 0 - 5
Factorise the quadratic:
(x + 2)(x - 1) = 0.
The solutions are x = -2 and x = 1.
- 6
The problem states that the x-coordinate of P is positive, so we must use x = 1.
- 7
To find the y-coordinate of P, substitute x = 1 back into the original equation for the curve:
y = 2(1)³ + 3(1)² - 12(1) + 1 - 8
Calculate the y-coordinate:
y = 2 + 3 - 12 + 1 = 5 - 12 + 1 = -6
Answer: -6
Common mistakes
- ×Confusing the value of the function, y, with the value of the gradient, dy/dx, at a specific point. Remember to substitute into the correct expression.
- ×Making arithmetic mistakes when evaluating the derivative at a point, especially with negative x-values.
- ×Substituting the x-value back into the derivative (dy/dx) instead of the original function (y) when asked to find the coordinates of a point on the curve.
No-calculator tips
- ✓Sketching the graph of a simple function (like a quadratic or cubic) can give you a visual clue about whether the gradient should be positive, negative or zero at a certain point, helping you to spot obvious errors.
- ✓When solving polynomial equations derived from setting the derivative to zero, always look for common factors to simplify the equation before attempting to factorise or solve.