If there is one prayer that you should

- Samuel Dominic Chukwuemeka
**pray/sing** every day and every hour, it is the
LORD's prayer (Our FATHER in Heaven prayer)

It is the **most powerful prayer**.
A **pure heart**, a **clean mind**, and a **clear conscience** is necessary for it.

For in GOD we live, and move, and have our being.

- Acts 17:28

The

- Samuel Dominic Chukwuemeka**Joy** of a **Teacher** is the **Success** of his **Students**.

I greet you this day,

__First:__ read the notes.

__Second:__ view the videos.

__Third:__ solve the solved examples and word problems

__Fourth:__ check your solutions with my **thoroughly-explained** solutions.

__Fifth:__ check your answers with the calculators as applicable.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.

If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting.

**Samuel Dominic Chukwuemeka** (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Students will:

(1.) Discuss the concept of the derivative of a function.

(2.) Determine the derivative of an explicit function by limit.

(3.) Determine the derivative of an explicit function by rules.

(4.) Determine the higher-order derivatives of explicit functions.

(5.) Differentiate implicit functions.

(6.) Differentiate exponential functions.

(7.) Differentiate logarithmic functions.

(8.) Differentiate trigonometric functions.

(9.) Solve applied problems involving the derivatives of functions.

**Pre-requisite Topic:** Exponents

The **Power Rule** of Derivatives states that:

$
If\:\: y = x^n \\[3ex]
then\:\: \dfrac{dy}{dx} = nx^{n - 1} \\[5ex]
If\:\: y = ax^n \\[3ex]
then\:\: \dfrac{dy}{dx} = nax^{n - 1}
$

The **Sum Rule** of Derivatives states that:

$
If\:\: y = u + v \\[3ex]
where\:\: u = f(x);\:\: v = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} \\[5ex]
If\:\: y = u + v + w \\[3ex]
where\:\: u = f(x);\:\: v = f(x);\:\: w = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} + \dfrac{dw}{dx}
$

The **Difference Rule** of Derivatives states that:

$
If\:\: y = u - v \\[3ex]
where\:\: u = f(x);\:\: v = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} - \dfrac{dv}{dx} \\[5ex]
If\:\: y = u - v - w \\[3ex]
where\:\: u = f(x);\:\: v = f(x);\:\: w = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} - \dfrac{dv}{dx} - \dfrac{dw}{dx}
$

You can write it in a compact form

The **Sum/Difference Rule** of Derivatives states that:

$
If\:\: y = u \pm v \\[3ex]
where\:\: u = f(x);\:\: v = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \\[5ex]
If\:\: y = u \pm v \pm w \\[3ex]
where\:\: u = f(x);\:\: v = f(x);\:\: w = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \pm \dfrac{dw}{dx}
$

The **Chain Rule** is also known as the **Function of a Function Rule**

We shall see the reasons.

Let us begin with the reason why it is called the Function of a Function Rule

$y$ is not a direct function of $x$

Rather, it is a function of some other function that is a function of $x$

The **Function of a Function Rule** of Derivatives states that:

$
If\:\: y = f(u) \\[3ex]
and\:\: u = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \\[5ex]
$
As you can see, $y$ is a function of $u$ which is then a function of $x$

So, $y$ is a function of a function of $x$

*
Student: Does this apply only to the case of three variables - $y, u, x$
where $y$ is a function of $u$ which is a function of $x$
Teacher: Good question.
No, it does not apply to only three variables.
It can apply to so many variables - sort of like a chain
Hence, the name "Chain Rule"
*

The **Chain Rule** of Derivatives states that:

$
If\:\: y = f(u) \\[3ex]
and\:\: u = f(v) \\[3ex]
and\:\: v = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dx} \\[5ex]
$
As you can see, this is a case of $3$ chains

$
If\:\: y = f(u) \\[3ex]
and\:\: u = f(v) \\[3ex]
and\:\: v = f(w) \\[3ex]
and\:\: w = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dw} * \dfrac{dw}{dx} \\[5ex]
$
This is a case of $4$ chains

*
Student: How do we set up the chains?
How do we know we set it up correctly?
How do we know when to use it?
Teacher: Good question.
When you see composite functions or functions with several exponents or functions that are not
"direct" sums, differences, product, or quotient; then the Chain Rule is probably the likely
rule to use
You do want to break it up into several functions as simply as possible
After setting it up and multiplying the functions, it will result to $\dfrac{dy}{dx}$
All the terms will cancel out leaving only the $\dfrac{dy}{dx}$
That way, you know you set it up correctly.
If you review the $2$ chains, $3$ chains, and $4$ chains, you will notice that only the
$\dfrac{dy}{dx}$ remains.
*

The **Product Rule** of Derivatives states that:

$
If\:\: y = u * v \\[3ex]
where\:\: u = f(x);\:\: v = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \\[5ex]
$
__NOTE:__ Before using the Product Rule, you must first express the function as the product of
**only two functions**.

In other words, it must be the **product of only two functions.**

**Pnuemonic to Remember Product Rule**

Let:
$
y = u * v \\[3ex]
u = first \\[3ex]
v = second \\[3ex]
y = first * second \\[3ex]
$
$\dfrac{dy}{dx} = first * dee-second + second * dee-first$

The **Quotient Rule** of Derivatives states that:

$
If\:\: y = \dfrac{u}{v} \\[5ex]
where\:\: u = f(x);\:\: v = f(x) \\[3ex]
then\:\: \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} \\[5ex]
$
__NOTE:__ Before using the Quotient Rule, you must first express the function as the quotient of
**only two functions**.

In other words, it must be the **quotient of only two functions.**

**Pnuemonic to Remember Quotient Rule**

Let:
$
y = \dfrac{u}{v} \\[5ex]
u = top \\[3ex]
v = bottom \\[3ex]
y = \dfrac{top}{bottom} \\[5ex]
$
$\dfrac{dy}{dx} = [(bottom * dee-top) \:\:minus\:\: (top * dee-bottom)] \:\:all\:\:over\:\: bottom-squared$

**Pre-requisite Topics:**

Limits and Continuity (Special Limits)

Trigonometric Identities

Trigonometric Formulas

Quotient Rule

**Pre-requisite Knowledge (from the Pre-requisite Topics):**

$
(1.)\:\: \displaystyle{\lim_{\theta \to 0}} \dfrac{\sin\theta}{\theta} = 1...Special\:\:Limit \\[5ex]
(2.)\:\: \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex]
(3.)\:\: \cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex]
(4.)\:\: \tan\theta = \dfrac{\sin\theta}{\cos\theta} ...Quotient\:\:Identity \\[5ex]
(5.)\:\: \sin^2 \theta + \cos^2 \theta = 1...Pythagorean\:\:Identity \\[3ex]
(6.)\:\: \sec\theta = \dfrac{1}{\cos\theta}...Reciprocal\:\:Identity \\[5ex]
(7.)\:\: \csc\theta = \dfrac{1}{\sin\theta}...Reciprocal\:\:Identity \\[5ex]
(8.)\:\: \cot\theta = \dfrac{\cos\theta}{\sin\theta}
$

The basic trigonometric functions are:

$
(1.)\:\: \sin x \\[3ex]
(2.)\:\: \cos x \\[3ex]
(3.)\:\: \tan x \\[3ex]
(4.)\:\: \csc x \\[3ex]
(5.)\:\: \sec x \\[3ex]
(6.)\:\: \cot x
$

**(1.) Derivative of** $\boldsymbol{\sin x}$

We shall use the **Special Limit for** $\displaystyle{\lim_{\theta \to 0}} \dfrac{\sin\theta}{\theta}$, **Derivatives by Limits**, and **Trigonometric Formulas (Sum-to-Product Formulas)**

Based on **Derivatives by Limits (Derivatives from First Principle)**

$
y = \sin x \\[3ex]
y + \Delta y = \sin (x + \Delta x) \\[3ex]
\Delta y = \sin (x + \Delta x) - y \\[3ex]
\Delta y = \sin (x + \Delta x) - \sin x \\[3ex]
\sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha + \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex]
Let\:\: \alpha = x + \Delta x \\[3ex]
Let\:\: \beta = x \\[3ex]
\rightarrow \sin (x + \Delta x) - \sin x = 2 \sin \left(\dfrac{x + \Delta x - x}{2}\right) \cos \left(\dfrac{x + \Delta x + x}{2}\right)...Sum-to-Product\:\:Formula \\[5ex]
\sin (x + \Delta x) - \sin x = 2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(\dfrac{2x + \Delta x}{2}\right) \\[5ex]
2x + \Delta x = 2\left(x + \dfrac{\Delta x}{2}\right) ...Factor\:\:by\:\:2 \\[5ex]
\dfrac{2x + \Delta x}{2} = \dfrac{2\left(x + \dfrac{\Delta x}{2}\right)}{2} = x + \dfrac{\Delta x}{2} \\[5ex]
\rightarrow \sin (x + \Delta x) - \sin x = 2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right) \\[5ex]
\rightarrow \Delta y = 2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right) \\[5ex]
Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex]
\dfrac{\Delta y}{\Delta x} = \dfrac{2 \sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right)}{\Delta x} \\[7ex]
Multiply\:\:both\:\:the\:\:numerator\:\:and\:\:denominator\:\:on\:\:the\:\:RHS\:\:by\:\:\dfrac{1}{2} \\[5ex]
\dfrac{2 * \dfrac{1}{2} * \sin \left(\dfrac{\Delta x}{2}\right) * \cos \left(x + \dfrac{\Delta x}{2}\right)}{\dfrac{1}{2} * \Delta x} = \dfrac{\sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex]
\rightarrow \dfrac{\Delta y}{\Delta x} = \dfrac{\sin \left(\dfrac{\Delta x}{2}\right) \cos \left(x + \dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex]
\dfrac{\Delta y}{\Delta x} = \dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} * \cos\left(x + \dfrac{\Delta x}{2}\right) \\[7ex]
Introduce\:\:limits \\[3ex]
\displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\theta \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} * \displaystyle{\lim_{\theta \to 0}}\cos \left(x + \dfrac{\Delta x}{2}\right) \\[7ex]
\displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \dfrac{dy}{dx} \\[5ex]
\displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{0}{2}\right)}{\dfrac{0}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin 0}{0} = 1 \\[7ex]
\displaystyle{\lim_{\Delta x \to 0}}\cos\left(x + \dfrac{\Delta x}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}\cos\left(x + \dfrac{0}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}\cos(x + 0) = \cos x \\[7ex]
\rightarrow \dfrac{dy}{dx} = 1 * \cos x \\[5ex]
\therefore \dfrac{dy}{dx} = \cos x
$

**(2.) Derivative of** $\boldsymbol{\cos x}$

We shall use the **Special Limit for** $\displaystyle{\lim_{\theta \to 0}} \dfrac{\sin\theta}{\theta}$, **Derivatives by Limits**, and **Trigonometric Formulas (Sum-to-Product Formulas)**

Based on **Derivatives by Limits (Derivatives from First Principle)**

$
y = \cos x \\[3ex]
y + \Delta y = \cos(x + \Delta x) \\[3ex]
\Delta y = \cos(x + \Delta x) - y \\[3ex]
\Delta y = \cos(x + \Delta x) - \cos x \\[3ex]
\cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha - \beta}{2}\right)...Sum-to-Product\:\:Formula \\[5ex]
Let\:\: \alpha = x + \Delta x \\[3ex]
Let\:\: \beta = x \\[3ex]
\rightarrow \cos(x + \Delta x) - \cos x = -2 \sin \left(\dfrac{x + \Delta x + x}{2}\right) \sin \left(\dfrac{x + \Delta x - x}{2}\right)...Sum-to-Product\:\:Formula \\[5ex]
\cos(x + \Delta x) - \cos x = -2 \sin \left(\dfrac{2x + \Delta x}{2}\right) \sin \left(\dfrac{\Delta x}{2}\right) \\[5ex]
2x + \Delta x = 2\left(x + \dfrac{\Delta x}{2}\right) ...Factor\:\:by\:\:2 \\[5ex]
\dfrac{2x + \Delta x}{2} = \dfrac{2\left(x + \dfrac{\Delta x}{2}\right)}{2} = x + \dfrac{\Delta x}{2} \\[5ex]
\rightarrow \cos(x + \Delta x) - \cos x = -2\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right) \\[5ex]
\rightarrow \Delta y = -2\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right) \\[5ex]
Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex]
\dfrac{\Delta y}{\Delta x} = \dfrac{-2\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\Delta x} \\[7ex]
Multiply\:\:both\:\:the\:\:numerator\:\:and\:\:denominator\:\:on\:\:the\:\:RHS\:\:by\:\:\dfrac{1}{2} \\[5ex]
\dfrac{-2 * \dfrac{1}{2} * \sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{1}{2} * \Delta x} = \dfrac{-\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex]
\rightarrow \dfrac{\Delta y}{\Delta x} = \dfrac{-\sin \left(x + \dfrac{\Delta x}{2}\right)\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex]
\dfrac{\Delta y}{\Delta x} = -\sin \left(x + \dfrac{\Delta x}{2}\right) * \dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex]
Introduce\:\:limits \\[3ex]
\displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\theta \to 0}}-\sin \left(x + \dfrac{\Delta x}{2}\right) * \displaystyle{\lim_{\theta \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} \\[7ex]
\displaystyle{\lim_{\Delta x \to 0}}\dfrac{\Delta y}{\Delta x} = \dfrac{dy}{dx} \\[5ex]
\displaystyle{\lim_{\Delta x \to 0}}-\sin\left(x + \dfrac{\Delta x}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}-\sin\left(x + \dfrac{0}{2}\right) = \displaystyle{\lim_{\Delta x \to 0}}-\sin(x + 0) = -\sin x \\[7ex]
\displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{\Delta x}{2}\right)}{\dfrac{\Delta x}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin \left(\dfrac{0}{2}\right)}{\dfrac{0}{2}} = \displaystyle{\lim_{\Delta x \to 0}}\dfrac{\sin 0}{0} = 1 \\[7ex]
\rightarrow \dfrac{dy}{dx} = -\sin x * 1 \\[5ex]
\therefore \dfrac{dy}{dx} = -\sin x
$

**(3.) Derivative of** $\boldsymbol{\tan x}$

We shall use the **Trigonometric Identities** and the **Quotient Rule**

$
y = \tan x = \dfrac{\sin x}{\cos x} ...Quotient\:\:Identity \\[5ex]
\dfrac{\sin x}{\cos x} = \dfrac{u}{v} \\[5ex]
u = \sin x \\[3ex]
\dfrac{du}{dx} = \cos x \\[5ex]
v = \cos x \\[3ex]
\dfrac{dv}{dx} = -\sin x \\[5ex]
\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex]
\dfrac{dy}{dx} = \dfrac{\cos x(\cos x) - \sin x(-\sin x)}{\cos^2x} \\[5ex]
\dfrac{dy}{dx} = \dfrac{\cos^2x + \sin^2x}{\cos^2x} \\[5ex]
\cos^2x + \sin^2x = 1 ...Pythagorean\:\:Identity \\[3ex]
\rightarrow \dfrac{dy}{dx} = \dfrac{1}{\cos^2x} \\[5ex]
\dfrac{1}{\cos x} = \sec x ...Reciprocal\:\:Identity \\[5ex]
\rightarrow \dfrac{1^2}{\cos^2x} = \dfrac{1}{\cos^2x} = \sec^2x \\[5ex]
\therefore \dfrac{dy}{dx} = \sec^2x
$

**(4.) Derivative of** $\boldsymbol{\csc x}$

We shall use the **Trigonometric Identities** and the **Quotient Rule**

$
y = \csc x = \dfrac{1}{\sin x} ...Reciprocal\:\:Identity \\[5ex]
\dfrac{1}{\sin x} = \dfrac{u}{v} \\[5ex]
u = 1 \\[3ex]
\dfrac{du}{dx} = 0 \\[5ex]
v = \sin x \\[3ex]
\dfrac{dv}{dx} = \cos x \\[5ex]
\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex]
\dfrac{dy}{dx} = \dfrac{\sin x(0) - 1(\cos x)}{\sin^2x} \\[5ex]
\dfrac{dy}{dx} = \dfrac{0 - \cos x}{\sin^2x} = \dfrac{-\cos x}{\sin^2x} = \dfrac{-\cos x}{\sin x} * \dfrac{1}{\sin x} \\[5ex]
\dfrac{-\cos x}{\sin x} = -\cot x ...Quotient\:\:Identity \\[5ex]
\dfrac{1}{\sin x} = \csc x ...Reciprocal\:\:Identity \\[5ex]
\rightarrow \dfrac{dy}{dx} = -\cot x * \csc x \\[5ex]
\therefore \dfrac{dy}{dx} = -\cot x\csc x
$

**(5.) Derivative of** $\boldsymbol{\sec x}$

We shall use the **Trigonometric Identities** and the **Quotient Rule**

$
y = \sec x = \dfrac{1}{\cos x} ...Reciprocal\:\:Identity \\[5ex]
\dfrac{1}{\cos x} = \dfrac{u}{v} \\[5ex]
u = 1 \\[3ex]
\dfrac{du}{dx} = 0 \\[5ex]
v = \cos x \\[3ex]
\dfrac{dv}{dx} = -\sin x \\[5ex]
\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex]
\dfrac{dy}{dx} = \dfrac{\cos x(0) - 1(-\sin x)}{\cos^2x} \\[5ex]
\dfrac{dy}{dx} = \dfrac{0 + \sin x}{\cos^2x} = \dfrac{\sin x}{\cos^2x} = \dfrac{\sin x}{\cos x} * \dfrac{1}{\cos x} \\[5ex]
\dfrac{\sin x}{\cos x} = \tan x ...Quotient\:\:Identity \\[5ex]
\dfrac{1}{\cos x} = \sec x ...Reciprocal\:\:Identity \\[5ex]
\rightarrow \dfrac{dy}{dx} = \tan x * \sec x \\[5ex]
\therefore \dfrac{dy}{dx} = \tan x\sec x
$

**(6.) Derivative of** $\boldsymbol{\cot x}$

We shall use the **Trigonometric Identities** and the **Quotient Rule**

$
y = \cot x = \dfrac{\cos x}{\sin x} ...Quotient\:\:Identity \\[5ex]
\dfrac{\cos x}{\sin x} = \dfrac{u}{v} \\[5ex]
u = \cos x \\[3ex]
\dfrac{du}{dx} = -\sin x \\[5ex]
v = \sin x \\[3ex]
\dfrac{dv}{dx} = \cos x \\[5ex]
\dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} ...Quotient\:\:Rule \\[5ex]
\dfrac{dy}{dx} = \dfrac{\sin x(-\sin x) - \cos x(\cos x)}{\sin^2x} \\[5ex]
\dfrac{dy}{dx} = \dfrac{-\sin^2x - \cos^2x}{\sin^2x} = \dfrac{-1(\sin^2x + \cos^2x)}{\sin^2x} \\[5ex]
\sin^2x + \cos^2x = 1 ...Pythagorean\:\:Identity \\[3ex]
\rightarrow \dfrac{dy}{dx} = \dfrac{-1}{\sin^2x} = -1 * \dfrac{1}{\sin^2x} \\[5ex]
\dfrac{1}{\sin x} = \csc x ...Reciprocal\:\:Identity \\[5ex]
\rightarrow \dfrac{1^2}{\sin^2x} = \dfrac{1}{\sin^2x} = \csc^2x \\[5ex]
\rightarrow \dfrac{dy}{dx} = -1 * \csc^2x \\[5ex]
\therefore \dfrac{dy}{dx} = -\csc^2x
$

**Implicit Differentiation** or **Implicit Differential Calculus** is the derivative of implicit
functions.

An **Implicit Function** is a function where the dependent and independent variables are expressed
**implicitly.** In other words, the dependent variable is not expressed in terms of the independent
variable.

In the previous examples we did - "Derivatives by Limits" and "Derivatives by Rules", we dealt with
explicit functions.

Here, we shall deal with implicit functions.

Say: $y=f(x)$

$y = x + 5 \rightarrow Explicit\:\: Function$

$y$ is expressed in terms of $x$

$y - x = 5 \rightarrow Implicit\:\: Function$

$y$ is not expressed in terms of $x$

*
Student: When we are given an implicit function, can we rearrange it to an explicit function
like in the example you just gave, and find the derivative using the Power Rule?
Teacher: Yes, you can.
However, the example is a very simple example.
We shall do more challenging ones as we progress through the course.
It will be better and less time-consuming to differentiate implicitly rather than trying to
convert it to an explicit function and differentiating explicitly.
Student: Do we have a whole new different set of rules for implicit differentiation?
It seems like it's a lot of rules.
Teacher: Good question. There are no new rules for implicit differentiation.
We shall learn a few more concepts/techniques and hence, have a broader understanding of the
concept of derivatives. But, there are no new rules for this section.
*

(1.) Differentiate each term with respect to the independent variable.

In other words, differentiate each term $wrt\:\: x$

(2.) Differentiating $x\:\:wrt\:\:x = \dfrac{dx}{dx} = 1$

(3.) Differentiating $x^2\:\:wrt\:\:x = 2x\dfrac{dx}{dx} = 2x * 1 = 2x$

(4.) Differentiating $x^3\:\:wrt\:\:x = 3x^2\dfrac{dx}{dx} = 3x^2 * 1 = 3x^2$

(5.) Differentiating $y\:\:wrt\:\:x = \dfrac{dy}{dx}$

(6.) Differentiating $y^2\:\:wrt\:\:x = 2y\dfrac{dy}{dx}$

(7.) Differentiating $y^3\:\:wrt\:\:x = 3y^2\dfrac{dy}{dx}$

(8.) Differentiating $p\:\:wrt\:\:x = \dfrac{dp}{dx}$

(9.) Differentiating $p^2\:\:wrt\:\:x = 2p\dfrac{dp}{dx}$

(10.) Differentiating $p^3\:\:wrt\:\:x = 3p^2\dfrac{dp}{dx}$

(11.) Differentiating

$
x^2y^3\:\:wrt\:\:x \\[3ex]
= x^2 * 3y^2\dfrac{dy}{dx} + y^3 * 2x\dfrac{dx}{dx} ...Product\:\:Rule \\[5ex]
= 3x^2y^2\dfrac{dy}{dx} + 2xy^3
$

(12.) Use the Rules of Derivatives as applicable.

**Vocabulary Words**

input, output, initial value, final value, change, increment, decrement, average rate of change, slope, secant line, intersects,
passes, tangent line, touches, quotient, difference quotient, limit, derivative, point of tangency,

*Recall:*

*Prior knowledge of PreAlgebra and Algebra*

**PreAlgebra:** (Relations and Functions):

(1.) We defined slope as: ratio of **the change in the output value** of the function with respect to (wrt)
**a unit change in the input value** of the function.

It is also known as the **average rate of change**

Considering a two-dimensional coordinate system where $y = f(x)$

$y$ = dependent variable

$x$ = independent variable

For a Linear Graph (Graph of a Linear Function) which is a straight line graph,

$
Point\;1\;\;(x_1, y_1) \\[3ex]
x_1 = initial\;\;value\;\;of\;\;x \\[3ex]
y_1 = initial\;\;value\;\;of\;\;y \\[3ex]
Point\;2\;\;(x_2, y_2) \\[3ex]
x_2 = final\;\;value\;\;of\;\;x \\[3ex]
y_2 = final\;\;value\;\;of\;\;y \\[3ex]
Change = final\;\;value - initial\;\;value \\[3ex]
Change\;\;in\;\;x = \Delta x = x_2 - x_1 \\[3ex]
Change\;\;in\;\;y = \Delta y = y_2 - y_1 \\[3ex]
Slope,\;\;m \\[3ex]
= \dfrac{\Delta y}{\Delta x} \\[5ex]
= \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex]
$
In this example, both changes are positive.

Hence, the slope is also positive.

As the value of $x$ increases, the value of $y$ also increases.

This implies that the rate of change of $y$ per unit change of $x$ is positive.

**Algebra:** (Difference Quotient):

(2.) We defined these terms:

(a.) The **secant line** to a curve is the **line that intersects two points** on the curve.

(b.) The

(

(3.) We reviewed the Difference Quotient of a function and noted these definitions.

We noted that:

(a.) The

(b.) The

(4.) We also noted that for:

Depending on the graph, the slope can be positive, negative, zero, or undefined.

But, how do we find the slope of non-linear graphs (curves)?

For example, how do we find the slope of a: parabola? graph of a cubic function? graph of a quartic function? etc.

Well, here comes Calculus!

Because a non-linear graph is not a straight line, the slope is not the same at every point on the graph/curve.

The slope changes/varies (rises, falls, is constant) from one point to another on the graph depending on the intervals at which the graph is increasing, decreasing, or is constant.

In this case, the slope of the curve is not the same for every point on the curve.

Let us look at the case of a parabola (the graph of a quadratic function).

The slope of the three points on the curve are different because the graph is not a straight line. It is a curve.

The slope of point C on the graph is negative because the graph is decreasing at certain intervals in which point C is located (red color)

The slope of point B is zero because the graph is constant at the interval that has point B (yellow color)

The slope of point A is positive because the graps is increasing at some intervals that has point A (green color)

Notice that there are three tangent lines to that curve because each line touches the curve at only one point: Point A or Point B or Point C

The points (Points A, B, and C) are known as points of tangency

The

So, the question is: how can we find the slope of any point say point A on a curve?

(1.)

This approach involves taking some measurements of the input ($x$) and the output ($y$) value that contains the point of tangency.

With this approach, it is important to note the scale of the graph on both axis: the $y-axis$ (horizontal axis) and the $x-axis$ (vertical axis)

$ Slope\;\;of\;\;the\;\;Curve\;\;at\;\;Point\;A \\[3ex] m = \dfrac{\Delta y}{\Delta x} \\[5ex] $ (2.)

This approach uses a secant line.

It is a more precise approach of approximating the slope of a curve at the point of tangency.

For a secant line, two points are needed.

The first point is the point of tangency.

The second point is another point on the curve.

Remember that we want to find the slope of the curve at Point A

Point A becomes out first point: Point 1

We find another point on the curve and call it Point 2

$ Slope\;\;of\;\;the\;\;Curve\;\;at\;\;Point\;A \\[3ex] Point\;A = Point\;1 = (x_1, y_1) \\[3ex] Point\;2 = (x_2, y_2) \\[3ex] y = f(x) \\[3ex] \Delta x = x_2 - x_1 \\[3ex] x_2 - x_1 = \Delta x \\[3ex] x_2 = \Delta x + x_1 \\[3ex] x_2 = x_1 + \Delta x \\[5ex] Let:\;\; x_1 = x \\[3ex] \implies \\[3ex] y_1 = f(x) \\[3ex] x_2 = x + \Delta x \\[3ex] y_2 = f(x_2) \\[3ex] y_2 = f(x + \Delta x) \\[3ex] Slope,\;\;m \\[3ex] = \dfrac{\Delta y}{\Delta x} \\[5ex] = \dfrac{y_2 - y_1}{x_2 - x_1} \\[5ex] = \dfrac{f(x + \Delta x) - f(x)}{(x + \Delta x) - x} \\[5ex] = \dfrac{f(x + \Delta x) - f(x)}{x + \Delta x - x} \\[5ex] = \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] Difference\;\;Quotient = DQ \\[3ex] DQ = \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] $ This is the

It is the

It is also known as the

The numerator is a difference.

The simplified denominator was a difference.

Here is the interesting thing:

The second point we chose is a bit distant.

We can choose a point very close to the first point (the point of tangency)

By choosing points closer and closer to the point of tangency (the first point), more accurate approximations of the slope of the tangent line is obtained.

This implies that as the secant line approaches the tangent line, more accurate approximations of the slope is obtained.

This implies that as the change in $x$: $\Delta x$ approaches zero, more accurate approximations of the slope is obtained.

Hence, the definition of limit is introduced.

(3.)

This approach uses the limit of the difference quotient as the change in the independent variable approaches zero.

It gives the exact value of the slope of the curve at the point of tangency.

$ \displaystyle{\lim_{\Delta x \to 0}} DQ \\[5ex] = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] = \dfrac{dy}{dx} \\[5ex] \therefore \dfrac{dy}{dx} = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] \dfrac{dy}{dx} = f'(x) \\[3ex] So: \\[3ex] f'(x) = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] $ This is the

It is the

It is also known as the

This gives us the exact value of the slope of the tangent line to the curve at Point A (the point of tangency)

It is the slope of the graph at $(x, f(x))$

Everything must not be $x$

We can use any variable as the independent variable.

Assume we decide to use the variable: $a$ rather than $x$, we have:

$ DQ = \dfrac{f(a + \Delta a) - f(a)}{\Delta a} \\[5ex] f'(a) = \displaystyle{\lim_{\Delta a \to 0}} \dfrac{f(a + \Delta a) - f(a)}{\Delta a} \\[5ex] $ The difference quotient is the approximate slope of the graph at $(a, f(a))$

The derivative is the exact slope of the graph at $(a, f(a))$

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