Differentiation
Section: Algebra and graphs 2 | Syllabus: Cambridge IGCSE Mathematics (0580)
Differentiating Power Functions
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Differentiation gives a formula for the gradient of a curve at any point, called the derived function or derivative. For a term of the form ax^n, there is a simple rule for finding its derivative.
- For a term ax^n, the derivative is anx^n-1 - multiply by the power, then reduce the power by 1
- For a sum of terms, differentiate each term separately and add the results
- A constant term (with no x) differentiates to 0, since a constant has zero gradient
- The derivative of y with respect to x is written (dy/dx)
Worked Example: Differentiating a Polynomial
- Question: Find (dy/dx) for y=4x^3-6x^2+5x-7.
- Step 1: Differentiate each term using anx^n-1: 4x^312x^2, -6x^2→-12x, 5x5, -70
- Step 2: Combine the results
- Answer: (dy/dx)=12x^2-12x+5
Common Mistakes
MistakeForgetting to reduce the power by 1 after multiplying, e.g. differentiating 4x³ as 12x³ instead of 12x²
Fixthe power rule has two parts - multiply by the original power, AND subtract 1 from the power
MistakeDifferentiating a constant term as if it were a variable term, e.g. leaving -7 unchanged
Fixa constant term always differentiates to 0 - it has no x, so its rate of change is zero
Finding the Gradient of a Curve at a Point
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Once the derivative is known, the gradient of the curve at any specific point can be found by substituting the x-coordinate of that point into the derivative.
- Differentiate the equation of the curve to find (dy/dx)
- Substitute the given x-value into (dy/dx) to find the gradient at that point
- The derivative itself is a function of x - it gives a different gradient value at each point on the curve
Worked Example: Finding the Gradient at a Point
- Question: A curve has equation y=x^3-4x^2+2. Find the gradient of the curve at x = 2.
- Step 1: Differentiate: (dy/dx)=3x^2-8x
- Step 2: Substitute x = 2: 3(2)^2-8(2)=12-16=-4
- Answer: gradient = -4
Common Mistakes
MistakeSubstituting the x-value into the original equation for y, instead of into the derivative, when asked for a gradient
Fixthe gradient always comes from dy/dx - substitute into the derivative, not the original equation for y
MistakeMaking an arithmetic slip when substituting a value into a squared term, e.g. computing 3(2)² as 3×2×2 in the wrong order
Fixwork out the power first (2²=4), then multiply by the coefficient
Finding Turning Points
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At a turning point, the gradient of the curve is momentarily zero - the curve is neither rising nor falling at that exact point.
- A turning point occurs where (dy/dx)=0
- To find the x-coordinate(s) of the turning point(s), differentiate the equation and solve (dy/dx)=0
- Substitute each x-value back into the original equation for y to find the corresponding y-coordinate
Worked Example: Finding a Turning Point
- Question: A curve has equation y=x^2-6x+5. Find the coordinates of its turning point.
- Step 1: Differentiate: (dy/dx)=2x-6
- Step 2: Set the derivative to 0: 2x-6=0 x=3
- Step 3: Substitute x = 3 into the original equation: y=(3)^2-6(3)+5=-4
- Answer: turning point at (3, -4)
Common Mistakes
MistakeSubstituting the x-value from dy/dx=0 back into the derivative instead of the original equation to find y
Fixthe y-coordinate of a turning point always comes from the original equation for y, not from dy/dx
MistakeStopping after finding only one solution to dy/dx=0 for a cubic, which can have two turning points
Fixdy/dx=0 for a cubic is a quadratic equation, which can have up to two solutions - solve it fully before concluding
Distinguishing Maximum and Minimum Points
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A turning point can be either a maximum or a minimum. Testing the gradient just before and just after the turning point shows which type it is.
- At a maximum point, the gradient changes from positive (rising) to negative (falling) as x increases through the turning point
- At a minimum point, the gradient changes from negative (falling) to positive (rising) as x increases through the turning point
- To test, substitute an x-value slightly less than, and slightly more than, the turning point's x-coordinate into dy/dx, and check the sign of each result
Worked Example: Testing a Turning Point
- Question: A curve has equation y=x^2-6x+5, with a turning point at x = 3. Determine whether this is a maximum or a minimum.
- Step 1: (dy/dx)=2x-6
- Step 2: Test x = 2 (slightly less than 3): 2(2)-6=-2 (negative)
- Step 3: Test x = 4 (slightly more than 3): 2(4)-6=2 (positive)
- Answer: the gradient changes from negative to positive, so this is a minimum
Common Mistakes
MistakeAssuming every turning point found is automatically a minimum, without testing
Fixalways test the gradient either side of the turning point before stating whether it is a maximum or minimum
MistakeChoosing test values that are too far from the turning point, potentially crossing into a different section of the curve
Fixpick test values close to the turning point's x-coordinate, one just below and one just above
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