Differentiation Basics and First Principles

The power rule is your best friend: for y = ax^n, the derivative is dy/dx = nax^$n-1$. Simply multiply by the power, then reduce the power by one.

Differentiation from first principles uses the formula f'(x) = lim(h→0) $f(x+h) - f(x)$/h. This shows you what's actually happening when you find a gradient. For f(x) = 3x², you substitute into the formula, expand f$x+h$, simplify by factoring out h, then let h approach zero.

The result? f'(x) = 6x, which matches what you'd get using the power rule. This method proves why the shortcuts work, though you'll mainly use it for exam questions asking you to "show from first principles."

Quick Check: Always verify your differentiation using the power rule - it's much faster than first principles!

Increasing and Decreasing Functions

Functions increase when their gradient is positive (f'(x) > 0) and decrease when it's negative (f'(x) < 0). To find these intervals, set f'(x) = 0 and solve for the boundary points.

For f(x) = x³ + 2x² - x + 2, you get f'(x) = 3x² + 2x - 1. Setting this equal to zero gives you x = 1/3 and x = -1. These are your critical points where the function changes behaviour.

Stationary points occur where f'(x) = 0 - these are your turning points and points of inflection. To determine their nature, use the second derivative test: f''(x) > 0 means minimum, f''(x) < 0 means maximum, and f''(x) = 0 suggests a point of inflection.

This method helps you sketch graphs and understand real-world problems like finding maximum profit or minimum cost.

Finding and Classifying Stationary Points

Let's work through f(x) = x²$3x² - 20$, which expands to f(x) = 3x⁵ - 20x³. The first derivative gives f'(x) = 15x⁴ - 60x², and setting this to zero reveals stationary points at x = 0, x = 2, and x = -2.

The second derivative f''(x) = 60x³ - 120x helps classify each point. At x = 2, f''(2) = 240 > 0, making (2, -64) a minimum point. At x = -2, f''(-2) = -240 < 0, making (-2, 64) a maximum point.

When f''(0) = 0, you've got a stationary point that needs further investigation - it could be a point of inflection. You'd need to check the sign of f'(x) on either side of x = 0 to determine its exact nature.

Remember to always find the y-coordinates by substituting back into the original function, not the derivative!

Pro Tip: Draw a sign diagram for f'(x) to visualise where your function increases and decreases between stationary points.

Differentiating Trigonometric and Exponential Functions

Trigonometric differentiation follows clear patterns: d/dx(sin x) = cos x, and d/dx(cos x) = -sin x. When you have d/dx(sin Rx) = R cos Rx, the coefficient R comes to the front - don't forget it!

Exponential functions have their own rules: d/dx(eˣ) = eˣ (it differentiates to itself), while d/dx(aˣ) = aˣ ln a. For logarithms, d/dx(ln x) = 1/x is essential to remember.

For composite functions like e^$-3x$, use d/dx$e^(Rx)$ = Re^(Rx). So e^$-3x$ becomes -3e^$-3x$. Similarly, 3^$x/2$ uses the rule d/dx$a^(Rx)$ = R(ln a)a^(Rx).

When finding gradients at specific points, substitute your x-value into the derivative. These rules are fundamental for many A-level applications, from population growth to wave motion.

Memory Aid: "eˣ is special - it stays the same when differentiated!"

The Chain Rule for Composite Functions

The chain rule tackles composite functions using dy/dx = $dy/du$ × $du/dx$. Think of it as differentiating from the outside in, then multiplying by the derivative of the inside function.

For y = e^(7x²), set u = 7x² (inside function) and y = eᵘ (outside function). Then dy/du = eᵘ and du/dx = 14x, giving dy/dx = eᵘ × 14x = 14xe^(7x²).

The same principle works for any composite function. With y = 5^(x⁴), you'd get dy/dx = (ln 5) × 5^(x⁴) × 4x³. The coefficient 4x³ comes from differentiating the inside function x⁴.

Practice identifying the "inside" and "outside" functions first - this makes the chain rule much clearer and prevents common mistakes.

Key Strategy: Always ask "what's inside the brackets or exponent?" - that's your u!

The Quotient Rule for Fractions

The quotient rule handles fractions: if y = u/v, then dy/dx = $v(du/dx) - u(dv/dx)$/v². Remember it as "bottom times top derivative minus top times bottom derivative, all over bottom squared."

For $x² + 1$/$x - 3$, set u = x² + 1 and v = x - 3. You get du/dx = 2x and dv/dx = 1, giving dy/dx = $(x-3)(2x) - (x²+1)(1)$/$x-3$² = $2x² - 6x - x² - 1$/$x-3$².

The quotient rule works brilliantly with trigonometric functions too. For sin x/cos x, you end up with $cos²x + sin²x$/cos²x = 1/cos²x, which is actually sec²x.

Sometimes you'll see mixed functions like cos(2x)/$x² + 2x + 1$. Just identify your u and v functions, differentiate each separately, then apply the formula methodically.

Remember: The order matters in the quotient rule - it's "bottom × top' - top × bottom'" in the numerator!

Simplifying Complex Quotient Rule Results

After applying the quotient rule, you'll often get messy expressions that need factoring and simplifying. For the function cos(2x)/$x² + 2x + 1$, notice that the denominator factors as $x + 1$².

The derivative becomes $(-2sin2x)(x+1)² - 2(cos2x)(x+1)$/$x+1$⁴. You can factor out $x+1$ from the numerator: $(x+1)[(-2sin2x)(x+1) - 2cos2x]$/$x+1$⁴.

This simplifies to $(-2sin2x)(x+1) - 2cos2x$/$x+1$³. Always look for common factors between numerator and denominator - it makes your final answer much cleaner.

These simplification skills are crucial for further calculus work and will save you time in exams. The key is recognising patterns and factoring opportunities before the expression gets too unwieldy.

Top Tip: Factor denominators first when you spot perfect squares or simple expressions - it often makes the quotient rule cleaner!