Equations of Tangents0%

Differentiation · Topic 8 of 15

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=dydxm = \frac{dy}{dx}), we can use this information directly with y−b=m(x−a)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: dydx\frac{dy}{dx} gives a formula for the gradient. You must substitute the xx-coordinate into it to get the number. Using the derivative itself as mm is a guaranteed loss.
  • Find the missing coordinate: if you are given only xx, substitute into the original equation — not the derivative — to get yy.
  • Tangent vs normal: a tangent uses mm; a normal uses −1m-\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=x2−3y = x^2 - 3 at the point (2, 1).

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

dydx=2x\frac{dy}{dx} = 2x

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

m=2(2)=4m = 2(2) = 4

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

y−b=m(x−a)y−1=4(x−2)y−1=4x−8y=4x−7\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 ff is defined on a suitable domain by f(x)=x3−2xf(x) = x^3 - 2x. Find the equation of the tangent to the curve y=f(x)y = f(x) when x=−2x = -2.

Step 1: We need the full coordinate. Find yy when x=−2x = -2.

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

Point is (-2, -4).

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

f′(x)=3x2−2m=f′(−2)=3(−2)2−2=3(4)−2=10\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=10m = 10 and (-2, -4):

y−(−4)=10(x−(−2))y+4=10(x+2)y+4=10x+20y=10x+16\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=x23y = \sqrt[3]{x^2} where x=−8x = -8.

Step 1: Find the full coordinate. Find yy when x=−8x = -8.

y=(x3)2=(−83)2=(−2)2=4\begin{aligned} y &= (\sqrt[3]{x})^2 \\ &= (\sqrt[3]{-8})^2 \\ &= (-2)^2 = 4 \end{aligned}

Point is (-8, 4).

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

y=x23dydx=23x−13m=23x3… when x=−8m=23(−83)=23(−2)=−26m=−13\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=−13m = -\frac{1}{3} and (-8, 4):

y−4=−13(x−(−8))y−4=−13(x+8)3(y−4)=−1(x+8)3y−12=−x−83y+x−4=0\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}