Gradient of a Curve and Equation of a Tangent (IGCSE Maths)
Finding the gradient of a curve at a point and the equation of a tangent are core IGCSE differentiation skills. This page shows you how to differentiate, substitute the x-value into dy/dx to get the gradient, and then use that gradient to write the full equation of a tangent to a curve in the form y = mx + c. Step-by-step worked examples cover every case, and the randomly generated, auto-marked practice questions let you test both skills instantly, with no sign-up needed and every answer checked automatically.
Be confident differentiating with the power rule before starting. Revise Differentiation (the power rule) and Gradient of a Straight Line first.
Move the point along the curve
Every point on the curve has its own gradient. The orange line is the tangent where the point sits, and its gradient is the value of \(\dfrac{dy}{dx}\) at that \(x\).
Finding the Gradient of a Curve at a Point
The gradient of a curve is different at every point. Differentiating gives you \(\dfrac{dy}{dx}\), the gradient function. To find the gradient at one specific point, substitute that \(x\)-value into \(\dfrac{dy}{dx}\). The answer is the gradient of the tangent to the curve there.
💡 Example 1: Gradient at a point
Find the gradient of \(y = 2x^3 - 5x\) at \(x = 3\).
\[\frac{dy}{dx} = 6x^2 - 5\]Substitute \(x = 3\):
\[ \begin{array}{rcl} \frac{dy}{dx} &=& 6(3)^2 - 5 \\ &=& 54 - 5 \\ &=& 49 \end{array} \]💡 Example 2: Negative power
Find the gradient of \(y = \dfrac{12}{x^2}\) at \(x = 2\).
Rewrite in index form: \(y = 12x^{-2}\)
\[\frac{dy}{dx} = -\frac{24}{x^3}\]Substitute \(x = 2\):
\[ \begin{array}{rcl} \frac{dy}{dx} &=& -\dfrac{24}{2^3} \\ &=& -\dfrac{24}{8} \\ &=& -3 \end{array} \]💡 Example 3: Expand the brackets first
Find the gradient of \(y = (2x - 1)(x + 3)\) at \(x = 1\).
Multiply out first, then differentiate:
\[y = 2x^2 + 5x - 3\] \[\frac{dy}{dx} = 4x + 5\]Substitute \(x = 1\):
\[ \begin{array}{rcl} \frac{dy}{dx} &=& 4(1) + 5 \\ &=& 9 \end{array} \]💡 Example 4: A root (fractional power)
Find the gradient of \(y = \sqrt{x}\) at \(x = 9\).
Rewrite as a power: \(y = x^{1/2}\), so \(\dfrac{dy}{dx} = \tfrac{1}{2}x^{-1/2}\).
\[\frac{dy}{dx} = \frac{1}{2\sqrt{x}}\]Substitute \(x = 9\):
\[ \begin{array}{rcl} \frac{dy}{dx} &=& \dfrac{1}{2\sqrt{9}} \\ &=& \dfrac{1}{6} \end{array} \]Equation of the Tangent to a Curve
A tangent is a straight line, so its equation has the form \(y = mx + c\). The gradient \(m\) comes from \(\dfrac{dy}{dx}\) at the given point, and you also need a point on the line, which comes from the curve itself.
A straight line that just touches the curve at one point. Its gradient equals \(\dfrac{dy}{dx}\) at that point.
The gradient \(m\) from differentiating, and a point \((x_1, y_1)\) on the curve.
\(y - y_1 = m(x - x_1)\), then rearrange into \(y = mx + c\).
⚡ Four Steps
- Substitute the given \(x\)-value into \(y\) to find the \(y\)-coordinate of the point.
- Differentiate to find \(\dfrac{dy}{dx}\).
- Substitute the \(x\)-value into \(\dfrac{dy}{dx}\) to get the gradient \(m\).
- Use \(y - y_1 = m(x - x_1)\) and rearrange to \(y = mx + c\).
💡 Example 5: Full worked example
Find the equation of the tangent to \(y = x^2 - 2x - 3\) at the point where \(x = 4\).
Step 1: find the point on the curve.
\[ \begin{array}{rcl} y &=& (4)^2 - 2(4) - 3 \\ &=& 16 - 8 - 3 \\ &=& 5 \end{array} \]So the point is \((4,\ 5)\).
Step 2: differentiate.
\[\frac{dy}{dx} = 2x - 2\]Step 3: gradient at \(x = 4\).
\[ \begin{array}{rcl} m &=& 2(4) - 2 \\ &=& 6 \end{array} \]Step 4: equation of the tangent.
\[ \begin{array}{rcl} y - 5 &=& 6(x - 4) \\ y &=& 6x - 24 + 5 \\ y &=& 6x - 19 \end{array} \]💡 Example 6: Negative gradient tangent
Find the equation of the tangent to \(y = x^2 - 3x + 5\) at \(x = 1\).
Point: at \(x = 1\), \(y = 1 - 3 + 5\), so \(y = 3\) and the point is \((1,\ 3)\).
Gradient function: \(\dfrac{dy}{dx} = 2x - 3\)
Gradient at \(x = 1\): \(m = 2 - 3\), so \(m = -1\).
Tangent:
\[ \begin{array}{rcl} y - 3 &=& -1(x - 1) \\ y &=& -x + 4 \end{array} \]💡 Example 7: Rearranged answer
Find the equation of the tangent to \(y = 4x^2\) at \(x = -1\), in the form \(y = mx + c\).
Point: at \(x = -1\), \(y = 4(-1)^2\), so \(y = 4\) and the point is \((-1,\ 4)\).
Gradient: \(\dfrac{dy}{dx} = 8x\), so \(m = 8(-1)\), giving \(m = -8\).
Tangent:
\[ \begin{array}{rcl} y - 4 &=& -8(x + 1) \\ y &=& -8x - 8 + 4 \\ y &=& -8x - 4 \end{array} \]🔑 Key Points
- Rewrite roots and fractions as powers of \(x\) before differentiating (\(\sqrt{x} = x^{1/2}\), \(\tfrac{k}{x^n} = kx^{-n}\)).
- Substitute \(x\) into \(\dfrac{dy}{dx}\) (not into \(y\)) to find the gradient.
- Multiply out brackets before differentiating; there is no product rule at this level.
- For a tangent you need both the gradient and a point: substitute \(x\) into \(y\) for the point.
- Give an exact fraction when the gradient is not a whole number (for example \(-\tfrac{5}{2}\)).
⚠️ Common Mistakes
- Using the \(y\)-value as the gradient instead of substituting into \(\dfrac{dy}{dx}\).
- Forgetting to find the \(y\)-coordinate; a tangent needs a full point, not just \(x\).
- Rounding the gradient too early and carrying an error into the final equation.
- Leaving the tangent as \(y - y_1 = m(x - x_1)\) when the question asks for \(y = mx + c\).
- Sign slips with negative powers: \(\dfrac{d}{dx}(x^{-2}) = -2x^{-3}\), not \(2x^{-3}\).