The derivative of $f(x) = 5^{x^2}$ with respect to $x$ is:

B) $2x \cdot 5^{x^2} \cdot \ln 5$

The sum of the first n terms of an arithmetic sequence (S_n) is formulated as:

S_n = n / 2 * [ 2a + (n - 1) * d ] = n / 2 * [ a + l ]

The sum to infinity (S_infinity) of a geometric series exists and converges if and only if the common ratio satisfies:

|r| < 1 (i.e. -1 < r < 1), and S_infinity = a / (1 - r)

كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

Implicit Differentiation, Parametric Differentiation and Higher Derivatives

Q: كيف يحسب الرقم الهيدروجيني pH للمحاليل؟ How to calculate pH and pOH?

يحسب الرقم الهيدروجيني بالعلاقة: pH = -log[H⁺]، وللمحاليل المائية عند 25°C: pH + pOH = 14. إذا كان pH 7 فالمحلول قاعدي. [الصيغة الرمزية الموزونة: pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴]

This lesson covers lessons 1-1 and 1-2 of the calculus unit "Differentiation and its Applications".
After it you should be able to:

differentiate a relation in $x$ and $y$ when $y$ cannot be isolated;
find $\frac{d y}{d x}$ for a curve given by a parameter, and the derivative of one function with respect to another;
find second, third and $n$ th derivatives, including for implicit and parametric curves.

🌳 Concept tree

Implicit Differentiation, Parametric Differentiation and Higher Derivatives
├── 1-1 (a) Explicit and implicit relations
│   ├── Explicit functions
│   └── Implicit relations
├── 1-1 (b) Implicit differentiation
│   ├── The method
│   ├── The chain-rule step
│   └── Important remarks
├── 1-1 (c) Parametric differentiation
│   ├── Parametric curves
│   └── The rule
├── 1-1 (d) Derivative of one function with respect to another
│   └── The idea
├── 1-2 (a) Higher-order derivatives
│   ├── Definition and notation
│   ├── A common confusion
│   ├── Patterns for the nth derivative
│   └── Proving relations between derivatives
├── 1-2 (b) Second derivative of implicit relations
│   └── Method
├── 1-2 (c) Second derivative of parametric curves
│   └── The rule
└── 1-2 (d) Meaning of the derivatives
    ├── Graph connection
    └── Summary table

01 1-1 (a) Explicit and implicit relations

Explicit functions

An explicit function gives $y$ directly in terms of $x$ , for example $y = 4 x^{3} - 5 x + 2$ or $y = \sin (2 x + 3)$ .
For such a function the derivative is found by the ordinary rules.

Implicit relations

An implicit relation links $x$ and $y$ in one equation $f (x, y) = 0$ , for example $x y + y - 4 = 0$ or $x^{2} + y^{2} = 9$ .
Sometimes $y$ can be isolated: $x y + y - 4 = 0$ gives $y = \frac{4}{x + 1}$ for $x \neq - 1$ .
Sometimes one equation hides several functions: $x^{2} + y^{2} = 9$ is a circle, which fails the vertical-line test, but it splits into two functions.
The upper half is $y = \sqrt{9 - x^{2}}$ (domain $[- 3, 3]$ , range $[0, 3]$ ) and the lower half is $y = - \sqrt{9 - x^{2}}$ (range $[- 3, 0]$ ).
Each half is differentiable for $x \in] - 3, 3 [$ .
In many equations, such as $x^{3} + y^{3} = 6 x y$ , isolating $y$ is impossible or very awkward, so we differentiate the equation as it stands.

02 1-1 (b) Implicit differentiation

The method

Differentiate both sides of the equation with respect to $x$ .
Treat $y$ as a differentiable function of $x$ and apply the chain rule to every term containing $y$ .
Collect all terms containing $\frac{d y}{d x}$ on one side, factorise, and divide.

The chain-rule step

$\frac{d}{d x} (y^{n}) = n y^{n - 1} \frac{d y}{d x}$ .
$\frac{d}{d x} (\sin y) = \cos y \frac{d y}{d x}$ and $\frac{d}{d x} (e^{y}) = e^{y} \frac{d y}{d x}$ .
$\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ (product rule).
If you differentiate with respect to $y$ instead, the factor $\frac{d x}{d y}$ appears.

✍️ Worked examples

Example 1: Find $\frac{d y}{d x}$ if $x^{2} + y^{2} = 25$ , then find the tangent and normal at $(3, 4)$ .

Solution:

Differentiate: $2 x + 2 y \frac{d y}{d x} = 0$ .
So $\frac{d y}{d x} = - \frac{x}{y}$ .
At $(3, 4)$ the slope is $- \frac{3}{4}$ .
Tangent: $y - 4 = - \frac{3}{4} (x - 3)$ , that is $3 x + 4 y = 25$ .
Normal slope is the negative reciprocal, $\frac{4}{3}$ , so the normal is $y - 4 = \frac{4}{3} (x - 3)$ .

Important remarks

The final formula usually contains both $x$ and $y$ .
To find the slope at a given $x$ you must first find the matching $y$ from the original equation.
A point must lie on the curve before you substitute it; check it in the equation.
A horizontal tangent needs the numerator of $\frac{d y}{d x}$ equal to zero; a vertical tangent needs the denominator equal to zero.

03 1-1 (c) Parametric differentiation

Parametric curves

If $x = f (t)$ and $y = g (t)$ with the same domain, the pair of equations describes one curve, and $t$ is the parameter.
Each value of $t$ gives one point $(x, y)$ of the curve.

The rule

If $f$ and $g$ are differentiable in $t$ and $\frac{d x}{d t} \neq 0$ , then $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$
This is just the chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ .
Horizontal tangent: $\frac{d y}{d t} = 0$ and $\frac{d x}{d t} \neq 0$ .
Vertical tangent: $\frac{d x}{d t} = 0$ and $\frac{d y}{d t} \neq 0$ .

✍️ Worked examples

Example 6: For $x = t^{2} + 1$ and $y = t^{3} - 3 t$ find $\frac{d y}{d x}$ when $t = 2$ .

Solution:

$\frac{d x}{d t} = 2 t$ and $\frac{d y}{d t} = 3 t^{2} - 3$ .
$\frac{d y}{d x} = \frac{3 t^{2} - 3}{2 t}$ .
At $t = 2$ : $\frac{12 - 3}{4} = \frac{9}{4}$ .

04 1-1 (d) Derivative of one function with respect to another

The idea

To differentiate $y = f (x)$ with respect to $z = g (x)$ , treat $x$ as the parameter of both. $\frac{d y}{d z} = \frac{d y / d x}{d z / d x}$

✍️ Worked examples

Example 9: Find the derivative of $4 x^{3} - 9 x^{2} + 5$ with respect to $3 x^{2} + 7$ .

Solution:

Let $y = 4 x^{3} - 9 x^{2} + 5$ and $z = 3 x^{2} + 7$ .
$\frac{d y}{d x} = 12 x^{2} - 18 x$ and $\frac{d z}{d x} = 6 x$ .
$\frac{d y}{d z} = \frac{12 x^{2} - 18 x}{6 x} = 2 x - 3$ .

05 1-2 (a) Higher-order derivatives

Definition and notation

If $y = f (x)$ is differentiable, then $y^{'} = \frac{d y}{d x} = f^{'} (x)$ is a new function.
If $y^{'}$ is differentiable, its derivative is the second derivative: $y^{''} = \frac{d^{2} y}{d x^{2}} = f^{''} (x)$ .
Repeating gives the third derivative $y^{'''} = \frac{d^{3} y}{d x^{3}}$ and, in general, the $n$ th derivative $y^{(n)} = \frac{d^{n} y}{d x^{n}} = f^{(n)} (x)$ , with $n$ a positive integer.
Derivatives from the second onwards are called higher derivatives.

A common confusion

$\frac{d^{2} y}{d x^{2}}$ means the derivative of $\frac{d y}{d x}$ .
${(\frac{d y}{d x})}^{2}$ means the square of the first derivative.
They are different quantities; never mix them.

✍️ Worked examples

Example 11: Find the third derivative of $y = x^{4} - 2 x^{2} + 5$ .

Solution:

$y^{'} = 4 x^{3} - 4 x$ .
$y^{''} = 12 x^{2} - 4$ .
$y^{'''} = 24 x$ .

Patterns for the $n$ th derivative

$y = x^{n}$ gives $y^{(n)} = n!$ , and every later derivative is $0$ .
$y = e^{k x}$ gives $y^{(n)} = k^{n} e^{k x}$ .
$y = \sin a x$ gives $y^{(n)} = a^{n} \sin (a x + \frac{n π}{2})$ ; for $y = \cos a x$ use $\cos (a x + \frac{n π}{2})$ .
Differentiating $\sin 2 x$ four times returns $16 \sin 2 x$ , so the pattern repeats every four steps.

Proving relations between derivatives

Differentiate the given relation as many times as the required relation demands, then eliminate constants.

✍️ Worked examples

Example 14: If $y = \tan x$ prove that $y^{''} = 2 y (1 + y^{2})$ .

Solution:

$y^{'} = \sec^{2} x = 1 + \tan^{2} x = 1 + y^{2}$ .
$y^{''} = 2 y y^{'} = 2 y (1 + y^{2})$ .

06 1-2 (b) Second derivative of implicit relations

Method

Find $y^{'}$ first, simplify it, then differentiate again.
Replace any $y^{'}$ that appears by its formula, and use the original equation to simplify.

✍️ Worked examples

Example 16: If $x^{2} + y^{2} = 4$ prove that $y^{3} y^{''} + 4 = 0$ .

Solution:

$2 x + 2 y y^{'} = 0$ , so $y^{'} = - \frac{x}{y}$ .
$y^{''} = - \frac{y - x y^{'}}{y^{2}} = - \frac{y + \frac{x^{2}}{y}}{y^{2}} = - \frac{y^{2} + x^{2}}{y^{3}}$ .
Since $x^{2} + y^{2} = 4$ , $y^{''} = - \frac{4}{y^{3}}$ , so $y^{3} y^{''} + 4 = 0$ .

07 1-2 (c) Second derivative of parametric curves

The rule

If $x = f (t)$ and $y = g (t)$ then $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$
First find $\frac{d y}{d x}$ as a function of $t$ , differentiate it with respect to $t$ , then divide by $\frac{d x}{d t}$ .
A frequent mistake is to write $\frac{d^{2} y / d t^{2}}{d^{2} x / d t^{2}}$ ; this is wrong.

✍️ Worked examples

Example 18: If $x = 2 n^{3} - 5$ and $y = 6 n^{2} + 1$ find $\frac{d^{2} y}{d x^{2}}$ at $n = 1$ .

Solution:

$\frac{d x}{d n} = 6 n^{2}$ and $\frac{d y}{d n} = 12 n$ .
$\frac{d y}{d x} = \frac{12 n}{6 n^{2}} = \frac{2}{n}$ .
$\frac{d}{d n} (\frac{2}{n}) = - \frac{2}{n^{2}}$ .
$\frac{d^{2} y}{d x^{2}} = \frac{- 2 / n^{2}}{6 n^{2}} = - \frac{1}{3 n^{4}}$ .
At $n = 1$ the value is $- \frac{1}{3}$ .

08 1-2 (d) Meaning of the derivatives

Graph connection

$f^{'} (x)$ is the slope of the tangent; $f^{''} (x)$ tells how that slope changes.
Where $f^{''} (x) > 0$ the slope is increasing, so the graph bends upwards like a cup; where $f^{''} (x) < 0$ the slope is decreasing and the graph bends downwards like a cap.
For a polynomial, each differentiation lowers the degree by one, so $f^{'}$ and $f^{''}$ have simpler graphs than $f$ .
Plotting $f$ , $f^{'}$ and $f^{''}$ with graphing software shows that $f^{'}$ crosses the $x$ -axis at the turning points of $f$ , and $f^{''}$ crosses it at the points where the bending changes.

Summary table

Point of comparison	Implicit differentiation	Parametric differentiation
Data given	One equation in $x$ and $y$	$x = f (t)$ , $y = g (t)$
First derivative	Differentiate both sides, solve for $y^{'}$	$\frac{d y / d t}{d x / d t}$
Answer contains	Usually $x$ and $y$	Only the parameter $t$
Second derivative	Differentiate $y^{'}$ again and substitute $y^{'}$	$\frac{d}{d t} (\frac{d y}{d x}) / \frac{d x}{d t}$

Implicit Differentiation, Parametric Differentiation and Higher Derivatives

🧭 Lesson points — pick a point and study it with its question bank

🌳 Concept tree

01 1-1 (a) Explicit and implicit relations

Explicit functions

Implicit relations

02 1-1 (b) Implicit differentiation

The method

The chain-rule step

Important remarks

03 1-1 (c) Parametric differentiation

Parametric curves

The rule

04 1-1 (d) Derivative of one function with respect to another

The idea

05 1-2 (a) Higher-order derivatives

Definition and notation

A common confusion

Patterns for the $n$ th derivative

Proving relations between derivatives

06 1-2 (b) Second derivative of implicit relations

Method

07 1-2 (c) Second derivative of parametric curves

The rule

08 1-2 (d) Meaning of the derivatives

Graph connection

Summary table

Comprehensive Curriculum Encyclopedia & Core Terminology — Implicit Differentiation, Parametric Differentiation and Higher Derivatives

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⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

🔍 In-Depth Explanations, Deductions & Core Concepts:

📊 Core Trigonometric and Reciprocal Derivative Formulas:

💡 High-Yield Exam Traps & Key Takeaways: