Teach Me Maths

Differentiation

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How to differentiate powers and algebraic expressions, find gradients and rates of change, and work out a second derivative.

What differentiation means

A straight line has the same gradient everywhere. A curve can be steep in one place, flat in another, and slope the opposite way somewhere else. Differentiation gives a rule for finding its gradient at any point where the derivative exists.

The gradient of a curve at a point is the gradient of its tangent there: the straight line with the same direction as the curve at that point.

If the curve has equation y=f(x), its gradient function is written \frac{dy}{dx} or f^{\prime}(x). It is also called the derivative. Put an x-value into this new function to find a gradient; put it into the original function to find a height.

The derivative also measures the rate of change of y with respect to x. For example, a gradient of 3 means that, near that point, a small increase in x produces about three times that increase in y

Explore the gradient

For y=x^2-2, the gradient function is \frac{dy}{dx}=2x. Move the point along the curve and watch the tangent turn.

Try it: move to x=-2, then x=0, then x=2. The gradients are -4, 0 and 4. A negative gradient slopes down from left to right; a positive gradient slopes up. At x=0, the tangent is horizontal.

The point at x=1 has height -1 but gradient 2. Height and gradient are different quantities.

The power rule

For a term kx^n, multiply the coefficient by the power, then subtract 1 from the power:

y=kx^n \quad\Longrightarrow\quad \frac{dy}{dx}=knx^{n-1}

In AQA Level 2 Further Maths, the powers you differentiate are integers, including negative integers. Differentiate a sum or difference one term at a time.

  • 5x^4 differentiates to 20x^3
  • -3x^2 differentiates to -6x
  • 7x differentiates to 7
  • A constant such as -9 differentiates to 0

A constant disappears because its graph is horizontal. Adding a constant moves a curve up or down without changing its gradients.

Worked example 1

Given y=4x^5-3x^2+7x-9, work out \frac{dy}{dx}

  1. Differentiate each term separately, keeping its sign.

  2. Your turn. Differentiate 4x^5. Enter the resulting term.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    4x^5 differentiates to 4\times5x^4=20x^4

  3. The next two terms give -6x and 7. The constant -9 gives 0
  4. Answer: \frac{dy}{dx}=20x^4-6x+7

Simplify before differentiating

The power rule works directly on a sum of powers of x. If the expression contains brackets, expand them first. If it is a fraction with a single power of x underneath, divide each term by that power first.

Do not differentiate the two factors separately and multiply the results. That does not give the derivative of their product. Expanding first puts the expression into a form you can differentiate term by term.

Worked example 2

Given y=2x^2(3x-4), work out \frac{dy}{dx}

  1. Your turn. Expand 2x^2(3x-4) before differentiating.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    y=6x^3-8x^2

  2. For the first term, 6\times3x^2=18x^2. For the second, -8\times2x=-16x
  3. Answer: \frac{dy}{dx}=18x^2-16x

Fractions and negative powers

A power of x in the denominator becomes a negative power:

\frac{1}{x}=x^{-1} \qquad \frac{1}{x^2}=x^{-2}

The same differentiation rule applies. For example, 5x^{-2} differentiates to -10x^{-3}, which is -\frac{10}{x^3}. Subtracting 1 from -2 gives -3

A numerical denominator stays part of the coefficient: \frac{x^3}{4}=\frac14x^3 differentiates to \frac34x^2. Do not move the 4 into the power.

Keep any restrictions from the original expression. An expression containing \frac{1}{x} or \frac{1}{x^2} is undefined at x=0

Worked example 3

Given y=\frac{6x^4-9x^2+12}{3x^2}, where x\ne0, work out \frac{dy}{dx}

  1. Divide every term in the numerator by 3x^2: y=2x^2-3+\frac{4}{x^2}
  2. Write the fraction as a negative power: y=2x^2-3+4x^{-2}

  3. Your turn. Differentiate 4x^{-2}. Enter the resulting term.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    4x^{-2} differentiates to 4\times(-2)x^{-3}=-8x^{-3}

  4. The other terms differentiate to 4x and 0
  5. Answer: \frac{dy}{dx}=4x-\frac{8}{x^3}, for x\ne0

Finding a gradient at a point

  • Differentiate the whole expression.
  • Substitute the given x-coordinate into the derivative.
  • Simplify to a number.

When a question asks for a gradient or rate of change, use \frac{dy}{dx}. Substituting into y gives a coordinate instead.

Worked example 4

The curve y=x^3+2x^2-5x+6 passes through (-2,16). Work out its gradient at this point.

  1. Differentiate: \frac{dy}{dx}=3x^2+4x-5
  2. Substitute x=-2: \frac{dy}{dx}=3(-2)^2+4(-2)-5

  3. Your turn. Work out 3(-2)^2+4(-2)-5

    3\times4-8-5=-1

  4. Answer: the gradient is -1. The curve slopes down from left to right at this point.

When the gradient is given

Sometimes the question gives a gradient and asks you to find an x-value or an unknown constant. Differentiate first, then set the derivative equal to the given gradient.

If the question asks for a point, find its y-coordinate using the original equation after you have found x

Worked example 5

The curve y=2x^3+kx^2-3x has gradient 11 when x=-1. Work out the constant k

  1. Treat k as a constant when differentiating: \frac{dy}{dx}=6x^2+2kx-3
  2. Use x=-1 and the given gradient: 6(-1)^2+2k(-1)-3=11
  3. Simplify: 3-2k=11

  4. Your turn. Solve 3-2k=11 for k

    -2k=8, so k=-4

  5. Answer: k=-4. Check: 6(-1)^2-8(-1)-3=11

Second derivatives

The second derivative, written \frac{d^2y}{dx^2}, is found by differentiating the first derivative again. It measures the rate of change of the gradient function.

If \frac{d^2y}{dx^2}>0 at a point, the gradient is increasing there. If it is negative, the gradient is decreasing. This is about how the gradient changes; it does not by itself tell you whether y is positive or whether the curve is rising.

The notation means differentiate twice. It does not mean square \frac{dy}{dx}. Second derivatives are also used to classify stationary points, covered in the Stationary Points topic.

Worked example 6

Given y=x^4-2x^3+5x, work out \frac{d^2y}{dx^2} and its value when x=2

  1. First differentiate once: \frac{dy}{dx}=4x^3-6x^2+5

  2. Your turn. Differentiate 4x^3-6x^2+5 again.

    Type powers with ^ and fractions with /, for example x^2 or (x+1)/3

    \frac{d^2y}{dx^2}=12x^2-12x

  3. At x=2, the second derivative is 12(2)^2-12(2)=24
  4. Answer: \frac{d^2y}{dx^2}=12x^2-12x, with value 24 at x=2. The gradient is increasing there.

Common mistakes

  • Reducing the power but forgetting to multiply by it: 4x^5 gives 20x^4, not 4x^4
  • Keeping a constant in the derivative. The derivative of -9 is 0
  • Treating a negative power as positive: subtracting 1 from -2 gives -3
  • Differentiating factors separately and multiplying. Expand the brackets first.
  • Substituting into the curve when asked for a gradient. Use the derivative.
  • Squaring the first derivative instead of differentiating it again.

Before you move on

You should be able to differentiate positive and negative integer powers, simplify brackets and fractions first, find a gradient at a point, use a given gradient to find an unknown, and differentiate twice.

Next, use a gradient and a point to find the equation of a line in Tangents and Normals.

Now try it: Differentiation practice questions

More on this topic: Differentiation worksheet with full solutions

Unofficial revision notes written by Teach Me Maths. Not produced or endorsed by AQA.