A-Level Maths 9709 - P3 Differentiation: Further Techniques, Higher Derivatives, Stationary Points

1. Further Differentiation Techniques

This section covers techniques for differentiating functions that don't directly fit the standard differentiation rules.

1.1 Implicit Differentiation

Used when the relationship between variables is not explicitly given as $y = f(x)$, but is defined implicitly, e.g., $x^2 + y^2 = 25$.

Steps:

1. Differentiate both sides of the equation with respect to $x$, remembering to use the chain rule where necessary.
2. Collect terms containing $dy/dx$ on one side of the equation.
3. Factor out $dy/dx$.
4. Solve for $dy/dx$.

Example: $x^2 + y^2 = 25$

Step	Equation	Differentiation
Differentiate both sides with respect to x	$x^2 + y^2 = 25$	$2x + 2y \frac{dy}{dx} = 0$
Collect terms with $\frac{dy}{dx}$	$2y \frac{dy}{dx} = -2x$
Solve for $\frac{dy}{dx}$	$\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y}$

1.2 Parametric Differentiation

Used when $x$ and $y$ are defined in terms of a parameter, say $t$, e.g., $x = \cos t$, $y = \sin t$.

Steps:

1. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
2. Use the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.

Example: $x = \cos t$, $y = \sin t$

Step	Equation	Differentiation
Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$	$x = \cos t$, $y = \sin t$	$\frac{dx}{dt} = -\sin t$, $\frac{dy}{dt} = \cos t$
Use the chain rule	$\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t$

1.3 Product Rule

For functions of the form $y = u(x)v(x)$.

Rule: $\frac{dy}{dx} = \frac{du}{dx}v(x) + u(x)\frac{dv}{dx}$.

1.4 Quotient Rule

For functions of the form $y = \frac{u(x)}{v(x)}$.

Rule: $\frac{dy}{dx} = \frac{v(x)\frac{du}{dx} - u(x)\frac{dv}{dx}}{v(x)^2}$.

1.5 Second Product Rule

For functions of the form $y = u(x)v(x)w(x)$.

Rule: $\frac{dy}{dx} = \frac{du}{dx}v(x)w(x) + u(x)\frac{dv}{dx}w(x) + u(x)v(x)\frac{dw}{dx}$.

1.6 Third Product Rule

For functions of the form $y = u(x)v(x)w(x)z(x)$.

Rule: $\frac{dy}{dx} = \frac{du}{dx}v(x)w(x)z(x) + u(x)\frac{dv}{dx}w(x)z(x) + u(x)v(x)\frac{dw}{dx}z(x) + u(x)v(x)w(x)\frac{dz}{dx}$.

1.7 Chain Rule (Extended)

Used when the composite function is more complex.

Example: $y = (3x^2 + 1)^{5}$

$\frac{dy}{dx} = 5(3x^2 + 1)^{4} \cdot (6x) = 30x(3x^2 + 1)^{4}$

2. Higher Derivatives

Derivatives of derivatives.

First derivative: $\frac{dy}{dx}$ or $y'$.

Second derivative: $\frac{d^2y}{dx^2}$ or $y''$.

Third derivative: $\frac{d^3y}{dx^3}$ or $y'''$.

And so on...

Example: $y = x^4 + 3x^2 - 5x + 2$

Derivative	Equation
First Derivative	$y' = 4x^3 + 6x - 5$
Second Derivative	$y'' = 12x^2 + 6$
Third Derivative	$y''' = 24x$

3. Stationary Points

Points where the derivative of a function is zero or undefined.

To find stationary points:

1. Find the first derivative, $y' = f'(x)$.
2. Set $y' = 0$ and solve for $x$. These are critical points.
3. Find the second derivative, $y'' = f''(x)$.
4. Evaluate $y''$ at each critical point.
• If $y'' > 0$, the point is a local minimum.
• If $y'' < 0$, the point is a local maximum.
• If $y'' = 0$, the test is inconclusive.

Example: $y = x^3 - 3x^2 + 1$

Step	Equation	Derivative	Value	Interpretation
Find the first derivative	$y = x^3 - 3x^2 + 1$	$y' = 3x^2 - 6x$
Set $y' = 0$ and solve for $x$	$3x^2 - 6x = 0 \Rightarrow 3x(x - 2) = 0$	$x = 0$ or $x = 2$
Find the second derivative	$y' = 3x^2 - 6x$	$y'' = 6x - 6$
Evaluate $y''$ at the critical points	At $x = 0$, $y'' = -6 < 0$ (local maximum)	At $x = 2$, $y'' = 6 > 0$ (local minimum)