Equations of Tangents

Video lesson · from 37:093 worked examples

One lesson video covers all of Differentiation, so it opens at 37:09 for this topic — not from the beginning.

Theory

Once we have determined the gradient of a curve at a particular point ( $m = \frac{dy}{dx}$ ), we can use this information directly with $y - b = m(x - a)$ to find the equation of the tangent line touching the curve at this point.

⚠️ Common Examiner Traps

Negative indices: candidates who can otherwise do this question routinely lose marks by differentiating negative indices incorrectly. Check that step before going on.
Gradient, then point: $\frac{dy}{dx}$ gives a formula for the gradient. You must substitute the $x$ -coordinate into it to get the number. Using the derivative itself as $m$ is a guaranteed loss.
Find the missing coordinate: if you are given only $x$ , substitute into the original equation — not the derivative — to get $y$ .
Tangent vs normal: a tangent uses $m$ ; a normal uses $-\frac{1}{m}$ . Read which one is asked for.

Worked examples

Example 1

Find the equation of the tangent to the curve with equation $y = x^2 - 3$ at the point (2, 1).

Step 1: Find the gradient ( $\frac{dy}{dx}$ )

\frac{dy}{dx} = 2x

Step 2: Evaluate the gradient at $x = 2$

m = 2(2) = 4

Step 3: Find the equation of the straight line using $m = 4$ and point $(a,b) = (2,1)$ :

\begin{aligned} y - b &= m(x - a) \\ y - 1 &= 4(x - 2) \\ y - 1 &= 4x - 8 \\ y &= 4x - 7 \end{aligned}

Example 2

A function $f$ is defined on a suitable domain by $f(x) = x^3 - 2x$ . Find the equation of the tangent to the curve $y = f(x)$ when $x = -2$ .

Step 1: We need the full coordinate. Find $y$ when $x = -2$ .

\begin{aligned} y &= (-2)^3 - 2(-2) \\ &= -8 + 4 = -4 \end{aligned}

Point is (-2, -4).

Step 2: Find gradient ( $f'(x)$ ) at $x = -2$ .

\begin{aligned} f'(x) &= 3x^2 - 2 \\ m &= f'(-2) = 3(-2)^2 - 2 \\ &= 3(4) - 2 = 10 \end{aligned}

Step 3: Equation using $m = 10$ and (-2, -4):

\begin{aligned} y - (-4) &= 10(x - (-2)) \\ y + 4 &= 10(x + 2) \\ y + 4 &= 10x + 20 \\ y &= 10x + 16 \end{aligned}

Example 3

Find the equation of the tangent to the curve with equation $y = \sqrt[3]{x^2}$ where $x = -8$ .

Step 1: Find the full coordinate. Find $y$ when $x = -8$ .

\begin{aligned} y &= (\sqrt[3]{x})^2 \\ &= (\sqrt[3]{-8})^2 \\ &= (-2)^2 = 4 \end{aligned}

Point is (-8, 4).

Step 2: Find gradient ( $\frac{dy}{dx}$ ) at $x = -8$ .

\begin{aligned} y &= x^{\frac{2}{3}} \\ \frac{dy}{dx} &= \frac{2}{3}x^{-\frac{1}{3}} \\ m &= \frac{2}{3\sqrt[3]{x}} \\ \dots \text{ when } x &= -8 \\ m &= \frac{2}{3(\sqrt[3]{-8})} \\ &= \frac{2}{3(-2)} \\ &= -\frac{2}{6} \\ m &= -\frac{1}{3} \end{aligned}

Step 3: Equation using $m = -\frac{1}{3}$ and (-8, 4):

\begin{aligned} y - 4 &= -\frac{1}{3}(x - (-8)) \\ y - 4 &= -\frac{1}{3}(x + 8) \\ 3(y - 4) &= -1(x + 8) \\ 3y - 12 &= -x - 8 \\ 3y + x - 4 &= 0 \end{aligned}