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.
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.
Gradient of the Curve at a Point
The orange line is the tangent at one point. Its gradient equals \(\dfrac{dy}{dx}\) at that \(x\)-value.
💡 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 \(2x^{-3}\).