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Differential Calculus

Differentiating turns a curve's equation into a formula for its steepness. With one rule you can find the gradient at any point, write the tangent line, and pin down the peaks and valleys of a graph.

MYP 5Extended MathsNumber Sequences & Pre-CalculusCriteria A · C · D~10 min read

Written by Fayek · MYP student

The power rule

The derivative of a function measures how fast its output changes, which on a graph is the gradient of the curve. We write it as \(\dfrac{dy}{dx}\), read "dee y by dee x". For any power of \(x\) there is one clean rule.

Power rule
To differentiate a power of \(x\), multiply by the power, then subtract 1 from it: \(\dfrac{d}{dx}(x^n) = n\,x^{\,n-1}\).
\[ \frac{d}{dx}\left(x^n\right) = n\,x^{\,n-1} \]

So \(\dfrac{d}{dx}(x^3) = 3x^2\), and \(\dfrac{d}{dx}(x) = 1\) (since \(x = x^1\) gives \(1 \cdot x^0 = 1\)). A constant on its own has gradient 0, because it never changes: \(\dfrac{d}{dx}(7) = 0\).

Differentiating polynomials

A polynomial is just a sum of powers, so differentiate it one term at a time. A number multiplying a power stays as a multiplier: \(\dfrac{d}{dx}(k\,x^n) = k\,n\,x^{\,n-1}\).

Worked example

Differentiate \(y = 3x^4 - 2x^2 + 5x - 7\).

1
Term \(3x^4\): multiply by 4, drop the power by 1, giving \(12x^3\).
2
Term \(-2x^2\): gives \(-2 \cdot 2 x^{1} = -4x\).
3
Term \(5x\): gives \(5\). The constant \(-7\) differentiates to \(0\).
\(\dfrac{dy}{dx} = 12x^3 - 4x + 5\).

Gradient and the tangent

The derivative is a formula for the gradient, so to find the steepness at a particular point just substitute that point's \(x\)-value. That gradient is also the slope of the tangent, the straight line that just touches the curve there. Combine it with the point using \(y - y_1 = m(x - x_1)\).

Worked example

Find the equation of the tangent to \(y = x^2 - 3x + 2\) at \(x = 4\).

1
Differentiate: \(\dfrac{dy}{dx} = 2x - 3\).
2
Gradient at \(x = 4\): \(m = 2(4) - 3 = 5\).
3
Find the \(y\)-coordinate: \(y = (4)^2 - 3(4) + 2 = 16 - 12 + 2 = 6\), so the point is \((4, 6)\).
4
Use \(y - y_1 = m(x - x_1)\): \(y - 6 = 5(x - 4)\), which expands to \(y = 5x - 20 + 6\).
Tangent: \(y = 5x - 14\).

Two coordinates, one gradient

A tangent line needs a point and a slope. Students often find the gradient but forget to work out the \(y\)-coordinate on the original curve. Always substitute back into \(y\), not into \(\tfrac{dy}{dx}\), for that point.

Stationary points

At the very top of a hill or bottom of a valley on a curve, the tangent is flat, so the gradient is zero. These are stationary points (also called turning points). Find them by setting \(\dfrac{dy}{dx} = 0\) and solving.

To classify each one, use the second derivative \(\dfrac{d^2y}{dx^2}\): if it is negative the point is a maximum, if positive it is a minimum.

Worked example

Find and classify the stationary points of \(y = x^3 - 3x^2 + 4\).

1
Differentiate: \(\dfrac{dy}{dx} = 3x^2 - 6x\).
2
Set it to zero and factor: \(3x(x - 2) = 0\), so \(x = 0\) or \(x = 2\).
3
Find each \(y\): at \(x = 0\), \(y = 4\); at \(x = 2\), \(y = 8 - 12 + 4 = 0\). Points are \((0, 4)\) and \((2, 0)\).
4
Second derivative \(\dfrac{d^2y}{dx^2} = 6x - 6\). At \(x = 0\) it is \(-6\) (maximum); at \(x = 2\) it is \(6\) (minimum).
Maximum at \((0, 4)\); minimum at \((2, 0)\).

Zero gradient, then classify

Setting \(\tfrac{dy}{dx} = 0\) locates the turning points; the second derivative tells you which is which. Laying out both stages clearly, the finding and the justifying, is what the marker credits under Criterion A (knowing), Criterion C (communicating), and Criterion D (applying).

Check yourself

1. Differentiate \(y = 5x^3 - 2x\). +

Term \(5x^3\) gives \(5 \cdot 3 x^2 = 15x^2\); term \(-2x\) gives \(-2\). So \(\dfrac{dy}{dx} = \mathbf{15x^2 - 2}\).

2. Find the gradient of \(y = x^2 + 2x\) at \(x = 3\). +

Differentiate: \(\dfrac{dy}{dx} = 2x + 2\). Substitute \(x = 3\): \(2(3) + 2 = \mathbf{8}\).

3. Find the stationary point of \(y = x^2 - 6x + 5\) and say whether it is a maximum or minimum. +

Differentiate: \(\dfrac{dy}{dx} = 2x - 6\). Set to zero: \(2x - 6 = 0\), so \(x = 3\). Then \(y = 9 - 18 + 5 = -4\), giving the point \((3, -4)\). The second derivative is \(2\), which is positive, so it is a minimum at \((3, -4)\).

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