Applications of Derivatives: Tangents, Normals, and Rates of Change

The Gradient of a Curve at a Point

The gradient of a tangent to a curve at a given point is found by evaluating the derivative of the function at that point. This is one of the most fundamental applications of differential calculus in the NSW HSC Mathematics Advanced course.

If y = f(x), then the derivative f'(x) (also written as dy/dx) gives a gradient function — a formula that outputs the instantaneous gradient at any value of x on the curve.

To find the gradient at a specific point, substitute the x-coordinate of that point into the derivative.

• Step 1: Differentiate f(x) to obtain f'(x).
• Step 2: Substitute the given x-value into f'(x).
• Step 3: The result is the gradient m of the tangent at that point.

Worked Example: Find the gradient of the tangent to y = x³ − 4x + 1 at the point where x = 2.

• Differentiate: dy/dx = 3x² − 4
• Substitute x = 2: dy/dx = 3(2)² − 4 = 3(4) − 4 = 12 − 4 = 8
• Verification: 3 × 4 = 12; 12 − 4 = 8. Confirmed.
• The gradient of the tangent at x = 2 is 8.

Note: the gradient of a curve at a point refers to the gradient of the tangent at that point — the two phrases are used interchangeably in HSC contexts.

Equation of the Tangent Line

Once the gradient m of the tangent is known, the equation of the tangent line can be found using the point-gradient formula:

y − y₁ = m(x − x₁)

where (x₁, y₁) is the point of tangency (the point on the curve where the tangent touches).

Worked Example: Find the equation of the tangent to y = x³ − 4x + 1 at the point where x = 2.

• Find the y-coordinate: Substitute x = 2 into the original curve:
  y = (2)³ − 4(2) + 1 = 8 − 8 + 1 = 1
  Verification: 8 − 8 = 0; 0 + 1 = 1. Confirmed. Point of tangency: (2, 1).
• Gradient: From the previous section, m = 8.
• Apply point-gradient formula:
  y − 1 = 8(x − 2)
  y − 1 = 8x − 16
  y = 8x − 15

The equation of the tangent is y = 8x − 15.

In some HSC questions the point of tangency is given directly as a coordinate pair, and you only need to verify it lies on the curve before proceeding. Always substitute back to check: at x = 2, y = 8(2) − 15 = 16 − 15 = 1. This matches the curve value, confirming the equation is correct.

Tangent lines are linear. Their equation is always in the form y = mx + b or equivalently ax + by + c = 0. The HSC typically accepts either form unless the question specifies otherwise.