Cubic Functions and Stationary Points

You'll often encounter cubic functions like f(x) = x³ - 6x² + 9x in your exams. The key is finding where the curve has maximum and minimum points by setting f'(x) = 0.

When you differentiate this function, you get f'(x) = 3x² - 12x + 9. Setting this equal to zero and solving gives you the x-coordinates of your stationary points. Remember that maximum points have f''(x) < 0, whilst minimum points have f''(x) > 0.

Quick Tip: Always check your stationary points using the second derivative test - it's the fastest way to determine their nature!

The graph shows a classic cubic shape with one maximum at point A and one minimum at point B(3, 0). This pattern appears frequently in exam questions, so get comfortable with identifying these features.

Derivative Graphs and Gradient Calculations

Understanding how to sketch derivative graphs from the original function is crucial for Paper 1. When the original function has a maximum, f'(x) crosses from positive to negative through zero. At minimum points, f'(x) crosses from negative to positive.

For gradient calculations, remember that finding f'(4) when f(x) = √x + 2/x² means differentiating first, then substituting. You'll get f'(x) = 1/(2√x) - 4/x³, so f'(4) = 1/4 - 4/64 = 1/4 - 1/16.

Tangent lines are another favourite exam topic. When the gradient equals a specific value $like 12 in the curve y = 6x² - x³$, you set the derivative equal to that value: 12x - 3x² = 12. Solve for x, then find the corresponding y-coordinate.

Remember: The derivative at any point gives you the gradient of the tangent line at that point.

Advanced Applications and Turning Points

Turning points questions often involve finding coordinates and determining their nature. For y = x³ - 3x² - 9x + 12, you'd differentiate to get y' = 3x² - 6x - 9, then solve 3x² - 6x - 9 = 0.

Composite functions like p(x) = f(g(x)) require the chain rule for differentiation. If f(x) = 3x + 1 and g(x) = x² - 2, then p(x) = 3$x² - 2$ + 1 and p'(x) = 6x.

Circle and parabola problems combine geometry with calculus. When a circle touches a parabola at two points, the tangent gradients at those points are parallel to the line joining the circle's centre to each point.

Pro Tip: Always expand and simplify your functions before differentiating - it makes the algebra much easier!

Maximum and minimum value problems on closed intervals require checking both stationary points and endpoints. For f(x) = x³ - 2x² - 4x + 6 on [0, 3], evaluate f at x = 0, x = 3, and any stationary points in between.

Complex Differentiation and Applications

Higher-order polynomials like y = x⁴ + 4x³ + 2x² - 20x + 3 can be tricky, but the method stays the same. Differentiate to find y' = 4x³ + 12x² + 4x - 20, then solve y' = 0. Sometimes you'll find only one stationary point, which you can verify using the discriminant.

Trigonometric differentiation appears in mechanics problems. When velocity v(t) = 8cos$2t - π/3$, acceleration a(t) = v'(t) = -16sin$2t - π/3$. The chain rule is essential here because of the $2t - π/3$ inside the cosine.

Fractional and root functions need careful handling. For f(x) = x√x - 3x - 2/(x√x), rewrite using indices first: f(x) = x^(3/2) - 3x - 2x^(-3/2). Then differentiate term by term.

Key Strategy: Always rewrite roots and fractions as powers before differentiating - it prevents silly mistakes.

For increasing functions, you need f'(x) > 0. After finding stationary points, test the sign of f'(x) in each interval to determine where the function is strictly increasing.

Final Tips and Common Mistakes

Practice makes perfect with differentiation past papers. Focus on recognising question types quickly, and always double-check your algebra when finding stationary points.