Differentiation

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Simple Differentiation:

$y = a x^{n} \Rightarrow \frac{d y}{d x} = a n x^{n - 1}$

Further Reading and Questions:

IGCSE Questions
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Level 2 Further Maths Questions [inc MS]
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IGCSE Past Paper Questions
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The Gradient Function
(or graphing the derivative)

Gradient function refers to the derivative of a function, represented as $d y / d x$ , which indicates the rate of change of the function with respect to $x$ at any point along the curve. Understanding how to sketch and interpret these functions is essential for solving problems related to motion, optimization, and curve analysis.

Gradient Function $d y / d x$
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Sketching Gradient Functions
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Understanding Sketching Derivatives - What is Sketching Derivatives?

Sketching derivatives involves creating a graph that represents the behaviour of a function's derivative. This helps in understanding how the original function behaves, including its increasing and decreasing intervals, as well as its local maxima and minima.

Key Steps in Sketching Derivatives:

Find the First Derivative:: Calculate the derivative of the function. This will help identify critical points where the function changes from increasing to decreasing or vice versa.
Determine Critical Points:: Set the first derivative equal to zero to find critical points. These points indicate where the slope of the function is zero.
Analyze Intervals:: Use a number line to test intervals around the critical points. Determine where the first derivative is positive (function is increasing) and where it is negative (function is decreasing).
Find Local Extrema:: Identify local maxima and minima based on the sign changes of the first derivative.
Second Derivative Test:: Calculate the second derivative to find points of inflection and determine concavity. This helps in sketching the curvature of the graph.

Example of Sketching a Derivative:

For a function $f (x) = \frac{x^{3}}{3} + x - \frac{2}{3}$ :

Step	Calculation/Result
First Derivative	$f^{'} (x) = - x^{2} + 1$
Critical Points	$f^{'} (x) = 0$ gives $x = ±1$
Increasing Intervals	(−∞,−1) and (1,∞)
Decreasing Intervals	(−1,1)
Second Derivative	$f^{''} (x) = - 2 x$
Points of Inflection	x=0

Conclusion:

By following these steps, you can effectively sketch the derivative of a function, gaining insights into its behaviour and characteristics. This process is essential for understanding calculus concepts and solving related problems.

Differentiation from First principles

$f^{'} = \underset{h \to 0}{Lim} (\frac{f (x + h) - f (x)}{h})$

Basic Derivatives

Function $f (x)$	Derivative $f^{'} (x)$
$a$ $(a \in ℝ)$	$0$
$x^{n}$ $(n \in ℝ)$	$n x^{n - 1}$
$e^{x}$	$e^{x}$
$ln x$ $(x > 0)$	$\frac{1}{x}$

Function $f (x)$	Derivative $f^{'} (x)$
$sin x$	$cos x$
$cos x$	$- sin x$
$tan x$	${sec}^{2} x$
$cosec x$	$- cosec x cot x$
$sec x$	$sec x tan x$
$cot x$	$- {cosec}^{2} x$
${sin}^{-1} x$	$\frac{1}{\sqrt{1 - x^{2}}}$
${cos}^{-1} x$	$\frac{-1}{\sqrt{1 - x^{2}}}$
${tan}^{-1} x$	$\frac{1}{1 + x^{2}}$

Function $f (x)$	Derivative $f^{'} (x)$
$cosh x$	$sinh x$
$sinh x$	$cosh x$
$tanh x$	${sech}^{2} x$
$cosech x$	$- cosech x coth x$
$sech x$	$- sech x tanh x$
$coth x$	$- {cosech}^{2} x$
${cosh}^{-1} x$	$\frac{1}{\sqrt{x^{2} - 1}}$
${sinh}^{-1} x$	$\frac{1}{\sqrt{x^{2} + 1}}$
${tanh}^{-1} x$	$\frac{1}{1 - x^{2}}$

Rules of Differentiation

Constant factor

\frac{d (a y)}{d x} = a \frac{d y}{d x}

Term wise

\frac{d}{d x} (f + g) = \frac{d f}{d x} + \frac{d g}{d x}

The Reciprocal Derivative

\frac{d y}{d x} = {(\frac{d x}{d y})}^{-1}

Composite Functions
and the Chain rule

y = y (u)

\Rightarrow

\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}

Factorised Expressions
and the Product rule

y = u v

\Rightarrow

\frac{d (u v)}{d x} = \frac{d u}{d x} v + u \frac{d v}{d x}

Algebraic Fractions
and the Quotient rule

y = \frac{u}{v}

\Rightarrow

f^{'} (\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}

where $y$ , $u$ and $v$ are functions of $x$

Implicit Differentiation

Differentiation Proofs

Chain Rule

\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}

Differentiation Chain Rule
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Differentiation Chain Rule
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Chain Rule
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Differentiation from First Principles

Differentiation from First Principles

$f^{'} = \underset{h \to 0}{Lim} (\frac{f (x + h) - f (x)}{h})$

Example:

Differentiate $y = x^{2} - 5$ using first priniciples

f^{'} = \underset{h \to 0}{Lim} (\frac{f (x + h) - f (x)}{h})

f^{'} = \underset{h \to 0}{Lim} (\frac{{(x + h)}^{2} - 5 - (x^{2} - 5)}{h})

f^{'} = \underset{h \to 0}{Lim} (\frac{x^{2} + 2 h x + h^{2} - 5 - x^{2} + 5}{h})

f^{'} = \underset{h \to 0}{Lim} (\frac{2 h x + h^{2}}{h})

f^{'} = \underset{h \to 0}{Lim} (2 x + h)

f^{'} = 2 x

Questions:

Differentiation from First Priniciples
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Past Paper Questions
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Differentiation from First Priniciples
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Analysing a Function with the Aid of Differentiation

Location:: $f (x) > 0$ where the function is above the $x$ -axis;; $f (x) < 0$ where the function is below the $x$ -axis;; $f (x) = 0$ gives the $x$ -intercept(s);; $f (0) = c$ gives the $y$ -intercept.
Increasing, Decreasing or Stationary:: $f^{'} (x) > 0$ where the function is increasing;; $f^{'} (x) < 0$ where the function is decreasing;; $f^{'} (x) = 0$ the locations of any stationary points.
Turning:: If $f^{"} (x) > 0$ then the function is convex;; If $f^{"} (x) < 0$ then the function is concave;; If $f^{"} (x) = 0$ then there is potentially a point of inflection (further testing required).
Stationary points:: If $f^{'} (x) = 0$ and $f^{"} (x) > 0$ then a local minimum point;; If $f^{'} (x) = 0$ and $f^{"} (x) < 0$ then a local maximum point;; If $f^{'} (x) = 0$ and $f^{"} (x) = 0$ then there is potentially a stationary point of inflection (further testing required).

Questions:

Increasing and Decreasing Functions Exam Questions
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Increasing/Decreasing Function Video:: Practice Questions [ MS ].
$d^{2} y / d x^{2}$ Video:: Practice Questions [ MS ].
Stationary Points Video:: Practice Questions [ MS ].
Application of Differentiation Video:: Practice Questions [ MS ].
Sketch curve knowing maxima/minima: Practice Questions [ MS ].
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