To find the equation of a tangent at a point:

1. Differentiate to find the gradient function \( \frac{dy}{dx} \).
2. Substitute the x-value into \( \frac{dy}{dx} \) to get the gradient \(m\).
3. Use the point and the gradient to form the line: \[ y - y_1 = m(x - x_1) \]

• Simple powers: \(x^3,\, 5x^4,\, -2x^{-3}\)
• Roots written as powers: \[ \sqrt{x} = x^{1/2}, \qquad \sqrt[3]{x} = x^{1/3} \]
• Reciprocals except \(x^{-1}\): \[ \frac{1}{x^2} = x^{-2}, \quad \frac{2}{x^3} = 2x^{-3} \]
• Multiples and sums of power functions: \[ y = 4x^3 + \frac{5}{x^2} - \sqrt{x} \]

Find the equation of the tangent to \(y = x^3\) at \(x = 2\).

Step 1: Differentiate. \[ y = x^3 \quad \Rightarrow \quad \frac{dy}{dx} = 3x^2 \] Step 2: Find gradient when \(x = 2\). \[ m = 3(2)^2 = 12 \] Step 3: Find the point on the curve. \[ y = 2^3 = 8 \Rightarrow (2,\,8) \] Step 4: Use \(y - y_1 = m(x - x_1)\). \[ y - 8 = 12(x - 2) \] Final tangent: \(y = 12x - 16\).

Find the tangent to \(y = \sqrt{x}\) at \(x = 4\).

Rewrite: \[ y = x^{1/2} \] Differentiate: \[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \] Gradient at \(x=4\): \[ m = \frac{1}{2}(4^{-1/2}) = \frac{1}{2}\cdot\frac{1}{2} = \frac{1}{4} \] Point on curve: \( y = \sqrt{4} = 2 \Rightarrow (4,2)\) Tangent: \[ y - 2 = \frac{1}{4}(x - 4) \]

Find the tangent to \(y = \frac{5}{x^3}\) at \(x = 1\).

Rewrite: \[ y = 5x^{-3} \] Differentiate: \[ \frac{dy}{dx} = 5(-3)x^{-4} = -15x^{-4} \] Gradient when \(x=1\): \[ m = -15(1)^{-4} = -15 \] Point: \(y = \frac{5}{1^3} = 5 \Rightarrow (1,5)\) Tangent: \[ y - 5 = -15(x - 1) \]

• Always rewrite roots and reciprocals as powers before differentiating.
• Never try to differentiate \(y = \frac{1}{x}\) at this stage.
• Use point-slope form: \(y - y_1 = m(x - x_1)\).
• Always calculate the point on the curve before writing the tangent.
• Show rewriting step — method marks are given for it.